Primera Práctica de Análisis Matemático II

1

UNIVERSIDAD NAIONAL DE HUANCAVELICAFACULTAD DE CIENCIAS DE INGENIERÍA

ESCUELA ACADÉMICO PROFESIONAL DE INGENIERÍA CIVIL - HUANCAVELICA

PRIMERA PRÁCTICA DE ANÁLISIS MATEMÁTICO II

I. DERIVADAS: 1. Encontrar la derivada de las funciones siguientes, usando la definición de derivada.

a) f ( x )=√x2+3 x−e2x …………………………………(Joe Johan Zúñiga Curipaco)

Solución:

f ' ( x )=limh→0

√( x+h )2+3 ( x+h )−e2 ( x+h )−√ x2+3 x−e2x

h

f ' ( x )=limh→0

√( x+h )2+3 ( x+h )−e2 ( x+h )−√ x2+3 x−e2x

h (√( x+h )2+3 ( x+h )−e2 ( x+h )+√x2+3 x−e2x

√( x+h )2+3 ( x+h )−e2 ( x+h )+√x2+3 x−e2x )f ' ( x )=lim

h→0

( x+h )2+3 ( x+h )−e2 ( x+h )−(x2+3x−e2x )h(√ (x+h )2+3 ( x+h )−e2 (x +h)+√ x2+3 x−e2x )

f ' ( x )=limh→0

x2+2 xh+h2+3 x+3h−e2x +2h−x2−3x+e2x

h (√ ( x+h )2+3 ( x+h )−e2 ( x+h )+√ x2+3x−e2 x)

f ' ( x )=limh→0

2xh+h2+3h−e2x +2h+e2 x

h(√ (x+h )2+3 ( x+h )−e2 (x +h)+√ x2+3 x−e2x )

2



f ' ( x )=limh→0

h (2 x+h+3 )−e2x (e2h−1 )h(√ (x+h )2+3 ( x+h )−e2 (x +h)+√ x2+3 x−e2x )

f ' ( x )=limh→0

h (2 x+h+3 )

h(√ (x+h )2+3 ( x+h )−e2 (x +h)+√ x2+3 x−e2x )−lim

h→0

e2x (e2h−1 )h (√ ( x+h )2+3 ( x+h )−e2 ( x+h )+√x2+3x−e2x)

f ' ( x )=limh→0

2 x+h+3

√( x+h )2+3 ( x+h )−e2 ( x+h )+√x2+3 x−e2x−lim

h→0

2e2x

(√( x+h )2+3 ( x+h )−e2 ( x+h )+√x2+3 x−e2x ) [ limh→0

(e2h−1 )2h ]

f ' ( x )=limh→0

2 x+h+3

√( x+h )2+3 ( x+h )−e2 ( x+h )+√x2+3 x−e2x−lim

h→0

2e2x

(√( x+h )2+3 ( x+h )−e2 ( x+h )+√x2+3 x−e2x )[ ln (e ) ]

f ' ( x )=limh→0

2 x+h+3

√( x+h )2+3 ( x+h )−e2 ( x+h )+√x2+3 x−e2x−lim

h→0

2e2x

(√( x+h )2+3 ( x+h )−e2 ( x+h )+√x2+3 x−e2x )

f ' ( x )= 2 x+32√x2+3 x−e2x

− 2e2x

2√ x2+3 x−e2x

3



f ' ( x )= 2 x+3−2e2x

2√x2+3 x−e2x

b) g ( x )=3√ x3−2x2+x …………………………………(Joe Johan Zúñiga Curipaco)

Solución:

g ' ( x )=limh→0

3√ ( x+h )3−2 ( x+h )2+( x+h )− 3√x3−2 x2+ xh

g' ( x )=limh→0

3√ ( x+h )3−2 ( x+h )2+( x+h )−3√x3−2 x2+ xh (

3√ (( x+h )3−2 ( x+h )2+( x+h ) )2+ 3√( x+h )3−2 ( x+h )2+( x+h ) 3√ x3−2x2+x+ 3√( x3−2x2+x )2

3√ (( x+h )3−2 ( x+h )2+( x+h ) )2+ 3√( x+h )3−2 ( x+h )2+( x+h ) 3√ x3−2x2+x+ 3√( x3−2x2+x )2 )

g ' ( x )=limh→0

( x+h )3−2 ( x+h )2+( x+h )− (x3−2x2+x )

h ( 3√ (( x+h )3−2 ( x+h )2+( x+h ) )2+ 3√( x+h )3−2 ( x+h )2+( x+h ) 3√ x3−2x2+x+ 3√ (x3−2x2+x )2)

g ' ( x )=limh→ 0

x3+3 x2h+3 xh2+h3−2h2−4 xh−2 x2+ x+h−x3+2x2−x


4



g ' ( x )=limh→0

3 x2h+3 x h2+h3−2h2−4 xh+h


g ' ( x )=limh→0

h (3x2+3 xh+h2−4 x+1 )


g ' ( x )=limh→ 0

3x2+3 xh+h2−4 x+1

( 3√ (( x+h )3−2 ( x+h )2+( x+h ) )2+ 3√( x+h )3−2 ( x+h )2+( x+h ) 3√ x3−2x2+x+ 3√ (x3−2x2+x )2)

g ' ( x )= 3 x2−4 x+13√(x3−2x2+x )2

5



c) h ( x )=5√2 x3−3x2+2 …………………………………(Joe Johan Zúñiga Curipaco)

Solución:

h ' ( x )=limh→0

5√2 ( x+h )3−3 ( x+h )2+2− 5√2 x3−3 x2+2h

A=2 ( x+h )3−3 ( x+h )2+2

B=2 x3−3 x2+2

h ' ( x )=limh→0

5√A−5√Bh (

5√A4+ 5√A3 5√B+5√A2 5√B2+ 5√A 5√B3+ 5√B4

5√A4+ 5√A3 5√B+5√A2 5√B2+ 5√A 5√B3+ 5√B4 )

h ' ( x )=limh→0

A−B

h ( 5√A4+ 5√ A3 5√B+5√A2 5√B2+ 5√ A 5√B3+ 5√B4 )

Remplazando valores:

h ' ( x )= limh→0

2 ( x+h )3−3 ( x+h )2+2−(2x3−3 x2+2 )h ( 5√A4+ 5√ A3 5√B+

5√A2 5√B2+ 5√ A 5√B3+ 5√B4 )

6



h ' ( x )=limh→0

2 x3+6 x2h+6 x h2+2h3−3 x2−6 xh−3h2+2−2 x3+3x2−2

h ( 5√A4+ 5√A3 5√B+5√A2 5√B2+ 5√A 5√B3+ 5√B4 )

h ' ( x )=limh→0

6x2h+6 x h2+2h3−6 xh−3h2

h ( 5√A4+ 5√ A3 5√B+5√A2 5√B2+ 5√ A 5√B3+ 5√B4 )

h ' ( x )= limh→0

h (6 x2+6 xh+2h2−6x−3h )h ( 5√A4+ 5√ A3 5√B+

5√A2 5√B2+ 5√ A 5√B3+ 5√B4 )

h ' ( x )= limh→0

(6 x2+6 xh+2h2−6x−3h )( 5√A4+ 5√ A3 5√B+

5√A2 5√B2+ 5√A 5√B3+ 5√B4 )

Reemplazando cuando h=0

A=2 ( x+h )3−3 ( x+h )2+2

A=2x3−3 x2+2

Decimos que A=B

h ' ( x )= 6 x2−6 x5√ (2x3−3 x2+2 )4+

5√(2 x3−3 x2+2 )4+5√ (2 x3−3 x2+2 )4+

5√ (2 x3−3x2+2 )4+5√(2x3−3 x2+2 )4

7



h ' ( x )= 6 x2−6 x

55√(2x3−3 x2+2 )4

d) i (x )=3√x4+3x−√x2+9 x

Solución:

i' ( x )=limh→0

3√ ( x+h )4+3 ( x+h )−√¿¿¿¿

i' ( x )= limh→0

¿¿

i' ( x )= limh→0

¿¿

i' ( x )=limh→0

¿¿

i' ( x )= limh→0

¿¿

8



i' ( x )=limh→0

¿¿

i' ( x )=limh→0

¿¿

i' ( x )=limh→0

¿¿

i' ( x )= 4 x3+3

[ 3√(x4+3 x )2+ 3√ (x4+3 x ) ( x4+3 x )+ 3√(x 4+3x )2]− 2 x+9

[√x2+9 x+√x2+9 x ]

i (x )= 4 x3+3

33√ (x4+3 x )2

− 2 x+9√ x2+9 x

Rpta

e) k ( x )=√cos (3 x )

k ' ( x )=limh→0

√cos (3 ( x+h ))−√cos (3 x)h

9



k ' ( x )=limh→0

√cos (3 ( x+h ))−√cos (3 x)h

√cos (3 ( x+h ))+√cos (3 x )√cos (3 ( x+h ))+√cos (3 x )

k ' ( x )=limh→0

cos (3 ( x+h ) )−cos (3x )

h1

√cos (3 ( x+h ))+√cos (3 x)

k ' ( x )=limh→0

cos (3 x )cos (3h )−sen (3 x ) sen(3h)−cos (3 x )

h

limh→0

1

√cos (3 ( x+h ))+√cos (3x )

k ' ( x )=[ limh→0 cos (3 x ) (cos (3h )−1 )

h−sen (3 x ) sen(3h)

h ][ 12√cos (3 x) ]

k ' ( x )=[−cos (3 x )limh→0

3 (1−cos (3h ) )

3h(1+cos (3h ) )(1+cos (3h ) )

−sen (3 x )limh→0

3 sen(3h)

3h ][ 12√cos (3 x) ]

k ' ( x )=[−3cos (3 x )limh→0

(1−cos2 (3h ) )3hx3h

1x 3h

(1+cos (3h ) )−3 sen (3x )(1)][ 1

2√cos (3 x) ]k ' ( x )=[−3cos (3 x ) lim

h→0 ( sen (3h)3h )

23 h

(1+cos (3h ) )−3 sen (3x )] [ 1

2√cos (3 x ) ]k ' ( x )=[−3cos (3 x ) lim

h→0 ( sen (3h)3h )

23 h

(1+cos (3h ) )−3 sen (3x )] [ 1

2√cos (3 x ) ]Por límites notables:

10



k ' (x)−3 sen (3 x )2√cos (3x )

Respuesta:

k ' (x)−3 sen (3 x )2√cos (3x )

f) l (x )=√sen ( x )

Solución:

l ' ( x )=limh→0

√sen ( x+h )−√sen (x )h

l ' ( x )=limh→0

√sen ( x+h )−√sen (x )h (√sen ( x+h )+√sen (x)

√sen ( x+h )+√sen (x))

l ' ( x )=limh→0

sen ( x+h )−sen ( x)

h ( 1

√sen ( x+h )+√sen (x ))l ' ( x )=

limh→0

sen ( x )cos (h)+sen (h ) cos (x)−sen (x)

h [ limh→0

1

√sen ( x+h )+√sen (x ) ]

11



l ' ( x )=[ limh→0sen ( x ) (cos (h )−1 )

h+limh→0

sen (h )cos (x )

h ] [ 12√sen (x) ]

l ' ( x )=[−sen ( x )limh→0

1−cos (h)

h+cos (x )

limh→0

sen (h )

h ] [ 12√sen (x) ]

l ' ( x )=[−sen ( x )limh→0

(1−cos (h))∗h

h∗h ( 1+cos (h)1+cos (h))+cos ( x)(1)][ 12√sen (x ) ]

l (x )=[−sen ( x ) limh→0 ( sen(h)h )

2

limh→0 ( h

1+cos (h))+cos (x)] [ 12√sen (x) ]

Por límites notables:

l ' ( x )=(−sen ( x ) (1 ) (0 )+cos (x ) ) [ 1

2√sen ( x ) ]l ' ( x )=[ cos ( x )

2√ sen ( x ) ]Respuesta:

12



l ' ( x )=[ cos ( x )2√ sen ( x ) ]

g) m ( x )=tan3( 2 x3 ) …………………………………(Eliseo Silvestre de la Cruz )

Solución:

¿limh→ 0

tan3( 2 ( x+h )3 )−tan3( 2x3 )h

limh→0

tan( 2 (x+h )3 )−tan( 2x3 ) [ tan2( 2 (x+h )

3 )−tan( 2 (x+h )3 ) tan( 2 x3 )+ tan

2

( 2 x3 )]h

limh→0

sen ( 2 (x+h )3 )

cos ( 2 ( x+h )3 )

−sen( 2 x3 )cos( 2 x3 )

hlimh→0 [ tan2( 2 ( x+h )

3 )−tan( 2 ( x+h )3 ) tan( 2 x3 )+tan

2

( 2 x3 )]

limh→0

sen( 2 ( x+h )3 )cos (2 x3 )−cos( 2 ( x+h )

3 )sen ( 2x3 )hcos ( 2 (x+h )

3 )cos( 2 x3 )[3 tan2( 2 x3 )]

( 23 limh→0sen (2 (h )

3 )2h3

) limh→0

1

cos ( 2 ( x+h )3 )cos ( 2 x3 ) [

3 tan2( 2 x3 )]

Por .limite notable:

13


ESCUELA ACADÉMICO PROFESIONAL DE INGENIERÍA CIVIL - HUANCAVELICA23 ( 1

cos2( 2x3 ) )[3 tan2(2x3 )]

m' ( x )=2 tan2( 2x3 )sec 2( 2 x3 )

h) n ( x )= ln4( 3 x2 ) …………………………………(Eliseo Silvestre de la Cruz )

Solución:

n'=limh→0

f ( x+h )−f (x )h

n'=limh→0

ln4 (3 x+3h2 )−ln4 (3 x2 )h

n'=limh→0

[( ln2( 3x+3h2 )+ ln2( 3 x2 ))( ln2( 3 x+3h2 )−ln2( 3 x2 ))]h

n'=limh→0

( ln2(3 x+3h2 )+ ln2( 3x2 ))( ln( 3x+3h2 )+ ln( 3 x2 ))(( ln( 3x+3h2 )−ln (3 x2 )))h

n'=limh→0 ( ln2( 3 x+3h2 )+ln2( 3 x2 ))( ln( 3 x+3h2 )+ln (3 x2 ))[ limh→0 ( ln( 3 x+3h2 )−ln( 3x2 ))

h ]n'=( ln2( 3 x2 )+ ln2( 3 x2 ))(ln( 3 x2 )+ ln(3 x2 ))[ 1h limh→0

ln( 3 x+3h3 x )]

14



n'=(2 ln2( 3 x2 ))(2 ln( 3x2 ))[ lim h→0

1hln( x+hx )]

n'=(4 ln3( 3 x2 )) [ limh→0 1h ln(1+ hx )]

n'=(4 ln3( 3 x2 )) ln(1+ hx )(xh ) 1x

n'=(4 ln3( 3 x2 )) ln (e )( 1x )

n'=(4 ln3( 3 x2 )) 1x ln (e )

n '=4 ln3( 3x2 ) 1x Rta.

i) ñ ( x )=log(√ x) …………………………………(Eliseo Silvestre de la Cruz )

Solución:

ñ '( x)=límh→0

log (√ x+h )−log(√ x)h

=límh→0

log √ x+hxh

ñ ' ( x )=log(límh→0(1+ hx )

12h)

ñ ' ( x )=log(límh→0(1+ hx )

1h12xx )=¿ log( límh→0(1+ hx )

xh)límh→0 ( 12x )¿

ñ ' ( x )=[ límh→0( 12x )] [ log (e) ]

15



ñ ' ( x )= log (e)2 x

j) q ( x )=ln (√3x )

Solución:

q ( x )=limh→0

ln (√3 ( x+h ) )−ln (√3 x)

h

q ' (x )=lim h→0

1hln(√ 3 ( x+h )

3 x )q ' (x )=lim

h→0ln( x+hx )

1h∙ 12

q ' (x )=ln limh→0 (1+ hx )

xh∙ 12x

q ' (x )=ln [elimh→0

1

2 x ]

q ' (x )=ln [elimh→0

1

2 x ]

q ' (x )= ln [ e 12 x ]

q ' (x )= ln [ e 12 x ]

q ' (x )= 12 x

( lne )

16



q ' ( x )= 12x

Respuesta:

q ' ( x )= 12x

k) r (x )=sec h2(x)………(Henry)

r ' (x )= 1

cosh2 ( x )= 1

( ex+e−x

2 )2= 4

e2 x+2+e−2x

r' (x )=lim

h→0

4

e2(x +h)+2+e−2 (x+h)− 4

e2x+2+e−2x

r ' (x )=

limh→0

4

e2 (x+h )+2+e−2 (x +h) −4

e2 x+2+e−2x

h

r ' (x )=

limh→0

4 (e2 x+2+e−2 x)−4 (e2(x+h)+2+e−2 ( x+h) )(e2 (x+h)+2+e−2 ( x+h ) ) (e2 x+2+e−2x )

h

r ' (x )=limh→0

4 (e2x+2+e−2x−e2 ( x+h )−2−e−2 ( x+h ) )

h (e2(x+ h)+2+e−2 ( x+h) ) (e2x+2+e−2x )

r ' (x )=limh→0

4 (e2x−e2 ( x+h)+e−2 x−e−2 (x +h) )

h (e2 (x+h )+2+e−2 (x +h) ) (e2x+2+e−2x )

r ' (x )=limh→0

4 (e2x (1−e2h)+e−2x (1−e−2h))

h (e2(x +h)+2+e−2 ( x+h )) (e2x+2+e−2 x)

r ' (x )=

limh→0

4 (e−2 x(1−1

e2h)+e2x (1−e

2h))h (e2 (x+h )+2+e−2 (x +h) ) (e2x+2+e−2x )

17



r ' (x )=

limh→0

4( e−2 x (e2h−1 )e2h

−e2 x (e2h−1))h (e2 (x+h )+2+e−2 (x +h) ) (e2x+2+e−2x )

r ' (x )=limh→0

4 (e2h−1) (e−2x−e2 xe2h )

h (e2 (x+h )+2+e−2 (x +h) ) (e2x+2+e−2x )e2h

r ' (x )=limh→0

4(eh+1)(eh−1)(e−2x−e2x e2h )

h (e2 (x+h )+2+e−2 (x +h) ) (e2x+2+e−2x )e2h

r ' (x )=[ limh→0( eh−1h )] [limh→0

(eh+1 )][ limh→0 ( 4

(e2 (x+h)+2+e−2( x+h) ) )] [ lim h→0( (e−2x−e2x e2h )

(e2 x+2+e−2 x)e2h )]r ' (x )=ln e .2( 2

e x+e−x )2

.(e−x−ex ) (e−x+ex )(ex+e− x) (ex+e−x )

r ' (x )=ln e .2( 2e x+e−x )

2

.¿

r ' (x )=−2 sech2 ( x ) tanh ( x )Rpta .

l) s ( x )=coth3 ( x )m) t ( x )=arcsen ( x )n) z (x )=arccos (x )o) w (x )=arctan ( x )

p) X(x)=arccot(x) ……… (henry)

SOLUCION:

limh→0

arccot ( x+h )−arccot(x )h

18



y '=1i

ln(√ x+ ix−i )

y '=limh→0

[ 1i ln(√ x+h+ix+h−i )−1i ln(√ x+i

x−i )]h

y '=limh→0

[ 1i ln(√ x+h+ix+h−i

√ x+ ix−i

)]h

y '=limh→0

[ 1i ln(√ ( x+h+i ) ( x+i )(x+h−i ) ( x−i ) )]

h

y '=limh→0

[ 1i ln(√ ( x+h+i ) ( x−i )(x+h−i ) ( x−i ) )]

h

y '=1i1hlimh→0

ln(√ ( x+h+i ) (x−i)( x+h−i ) ( x−i ) )

y '=1i1hlimh→0

ln(√ x2+xh+ ix−ix−ih+1x2+ xh+ix−ix+ ih+1 )

y '=1i1hlimh→0

ln(√ x2+xh−ih+1x2+xh+ih+1 )

y '=1iln [ limh→0 ( x2+xh−ih+1

x2+xh+ih+1 )12h ]

y '=1iln [ limh→0 (1+ x2+ xh−ih+1

x2+xh+ih+1−1)

12h ]

y '=1iln [ limh→0 (1+ x2+ xh−ih+1−x2−xh−ih−1

x2+xh+ih+1 )12h ]

y '=1iln [ limh→0 (1+ −2ih

x2+ xh+ih+1 )12h]

Llevando a la forma :

19



y '=1iln [ limh→0 (1+ −2ih

x2+ xh+ih+1 )12h(i (x2+ xh+ih+1 )

(x2+ xh+ih+1 ) )]y '=

1iln [ limh→0 (1+ −2ih

x2+ xh+ih+1 )12h(i (x2+ xh+ih+1 )

(x2+ xh+ih+1 ) ) limh→0 −i(x2+xh+ih+1) ]

y '=1iln e

limh→0 ( −i

(x2+xh+ih+1 ) )

y '=1ilimh→0 ( −i

(x2+xh+ih+1 ) ) ln eReemplazando (h→0 )

y '= −1x2+1

q) y ( x )=arcsec ( x )

r) z (x )=arccsc ( x ) (henry)

Solución:

csc ( y )=x

2 i

eiy−e−iy=x

2 i=x (e iy− 1

e iy ) 2 ie iy=xe2 iy−x

0=xe2 iy−2i eiy−x

0=x (eiy )2−2 i (e iy )−x

y=−b±√b2−4ac2a

e iy=2 i±√ (2i )2−4 ( x ) (−x )

2x

e iy=2 i±√4 i2+4 ( x ) ( x )

2x

20



e iy=i ±√x2−1x

lneiy=ln( i ±√ x2−1x )

iylne=ln( i ±√x2−1x )

y=1iln( i+√x2−1

x )

z ' (x)=lim

h→0

1iln( i+√( x+h )2−1

x+h )−1i ln( i+√ x2−1x )

h

z ' (x)=lim

h→0

1i [ ln( i+√ (x+h )2−1

x+h )−ln( i+√x2−1x )]

h

z ' (x)=limh→0

1i [ ln( i+√( x+h )2−1

x+hi+√ x2−1

x)]

h

z ' (x)=limh→0

1

i[ ln( i+√( x+h )2−1

x+hi+√ x2−1

x)

h]

z ' (x)=limh→0

1

ih [ ln( i+√ (x+h )2−1x+h

i+√x2−1x

)] z ' (x)=

limh→0

1

ih [ ln ( ( x ) (i+√ ( x+h )2−1)(x+h ) ( i+√x2−1 ) )]

21



z ' (x)=limh→0

1

ih [ ln ( xi+x√ ( x+h )2−1ix+ x√ x2−1+ ih+h√ x2−1 )]

z ' (x)=limh→0

1

ih [ ln ( xi+x √( x+h )2−1+ ix+x √x2−1+ih+h√x2−1−(ix+x √x2−1+ih+h√ x2−1 )ix+x √x2−1+ih+h√x2−1 )]

z ' (x)=limh→0

1

ih [ ln ( xi+x √( x+h )2−1+ ix+x √x2−1+ih+h√x2−1−ix−x √x2−1−ih−h√x2−1ix+x √x2−1+ih+h√ x2−1 )]

z ' (x)=limh→0

1

ih [ ln ( ( xi+x √ x2−1+ ih+h√ x2−1 )+ix+x √ ( x+h )2−1−ix−x √x2−1−ih−h√x2−1ix+ x√ x2−1+ih+h√x2−1 )]

z ' (x)=limh→0

1

ih [ ln (1+ x √ (x+h )2−1−x √x2−1−ih−h√x2−1ix+x √x2−1+ih+h√ x2−1 ) ]

z ' (x)=limh→0

1

ih [ ln (1+ x √ (x+h )2−1−x √x2−1−ih−h√x2−1ix+x √x2−1+ih+h√ x2−1 ) ]

(xi+ x √x2−1+ih+h√x2−1)(x √( x+h)2−1−x √x2−1−ih−h√x2−1)

(x √ ( x+h )2−1− x√ x2−1−ih−h√x2−1)(xi+ x√ x2−1+ ih+h√x2−1)

PROPIEDAD: ln e=ln(1+ 1h )h

z ' (x)=limh→0

1hi [ (x √( x+h )2−1−x √x2−1−ih−h√x2−1)

(xi+x √x2−1+ih+h√ x2−1 ) ] [ ln(1+ x√ ( x+h )2−1−x √x2−1−ih−h√ x2−1ix+x √ x2−1+ ih+h√ x2−1 )

(xi+ x √x2−1+ih+h√x2−1)(x √( x+h)2−1−x √x2−1−ih−h√x2−1) ]

z ' ( x )= limh→0

1hi [ ( x √( x+h )2−1−x√ x2−1−ih−h√ x2−1)

(xi+x √x2−1+ih+h√ x2−1 ) ] . lne z ' ( x )=lim

h→0

1hi [ x √( x+h )2−1−x √x2−1

xi+x √x2−1+ih+h√ x2−1− ih+h√ x2−1xi+x √x2−1+ih+h√ x2−1 ]

z ' ( x )=limh→0

1hi [ x √( x+h )2−1−x √x2−1

xi+x √x2−1+ih+h√ x2−1− ih+h√ x2−1xi+x √x2−1+ih+h√ x2−1 ]

22



z ' ( x )= limh→0

1hi [ x (√ ( x+h )2−1−√x2−1) (√ ( x+h )2−1+√ x2−1)

( xi+ x√ x2−1+ih+h√x2−1 ) (√( x+h )2−1+√x2−1)− ih+h√ x2−1xi+x √x2−1+ih+h√ x2−1 ]

z ' ( x )= limh→0

1hi [ x ( ( x+h )2−1− (x2−1 ))

( xi+ x√ x2−1+ih+h√x2−1 ) (√( x+h )2−1+√x2−1)− ih+h√ x2−1xi+x √x2−1+ih+h√ x2−1 ]

z ' ( x )=limh→0

1hi [ x ( ( x+h )2−1− (x2−1 ))

( xi+ x√ x2−1+ih+h√x2−1 ) (√( x+h )2−1+√x2−1) ]−[( 1hi )( ih+h√ x2−1xi+x √x2−1+ih+h√ x2−1 )]

z ' ( x )=limh→0

1hi [ x (x2+2 xh+h2−1−x2+1 )

( xi+ x√ x2−1+ih+h√x2−1 ) (√( x+h )2−1+√x2−1) ]−[( 1hi )( ih+h√ x2−1xi+x √x2−1+ih+h√ x2−1 )]

z ' ( x )=limh→0 [( 1hi )( 2 x2h+x h2

( xi+ x√ x2−1+ ih+h√ x2−1 ) (√( x+h )2−1+√x2−1) )]−[( 1hi )( ih+h√x2−1i+x√ x2−1+ih+h√x2−1 )]

z ' ( x )=limh→0 [(1i )( 2x2+xh

( xi+x √ x2−1+ ih+h√ x2−1 ) (√( x+h )2−1+√x2−1) )]−[( 1i )( i+√x2−1i+ x√ x2−1+ih+h√x2−1 )]

z' ( x )=[( 1i )( 2 x2

(xi+x √x2−1 ) (√x2−1+√ x2−1 ) )]−[(1i )( i+√x2−1i+x √x2−1 )]

z' ( x )= 1

(xi+x √x2−1 ) (1i )( 2 x2

2√x2−1−i−√x2−1)

z ' ( x )= 1

(xi+x √x2−1 ) (1i )( x

2−i √x2−1−x2+1

√x2−1 ) z ' ( x )= 1

(xi+x √x2−1 ) (1i )(−i .ix√ x2−1+ix

√x2−1 ) 1ix

23



z ' ( x )= 1

(xi+x √x2−1 ) ( x √x2−1+ix√x2−1 ) 1ix .i

z' ( x )= −1

x √x2−1Rpta .

s) A ( x )=arcsenh ( x )

t ¿ B (x )=arccosh ( x )…………….. (Joseph Michael Cayllahua Ch.)

solucion :

como sabemos ;cosx= ex+e− x

2

y=arccosh ( x )x=cosh ( y )

x=cosh ( y )= e y+e−iy

2⇒

ey+ 1e y

2

x= e y+12e y ⇒e2 y−2 xe y+1=0

e y=−(−2 x )±√ (−2x )2−4 (1 ) (1 )

2 (1 )

e y=2x ±√4 x2−42

⇒ x ±√x2+1

e y=x+√ x2−1 e y>0 ; x<¿ √ x2−1

e y=x−√x2−1

⇒ el signomenor sedescarta

e y=x+√ x2−1

y=ln (x+√x2−1 )

B' (x)=límh→0

f ( x+h )−f (x )h

B' (x)=límh→0

ln (( x+h )+√( x+h )2−1)−ln (x+√x2−1)h

24



B' (x)=límh→0

ln|( x+h )+√ ( x+h )2−1(x+√ x2−1 ) |

h

B'(x )=lím

h→0

1hln|( x+h )+√ ( x+h )2−1

(x+√ x2−1 ) |B'

(x )=límh→0

1hln|1+ ( x+h )+√ ( x+h )2−1− x−√x2−1

(x+√ x2−1 ) |ln|1+ ab|

abba⇒ a

bln|1+ a

b|ba=a

b(1 )

límh→0

1hln|1+ ( x+h )+√ ( x+h )2−1−x−√ x2−1

( x+√x2−1 ) |(( x+h)+√ ( x+h )2−1−x−√x2−1

(x+√x2−1 ) )( (x+√ x2−1 )( x+h)+√ ( x+h )2−1− x−√x2−1)

límh→0

1h ( (x+h )+√( x+h )2−1−x−√ x2−1

(x+√x2−1 ) ) ln|1+ ( x+h )+√ (x+h )2−1−x−√x2−1(x+√ x2−1 ) |(

(x+√x2−1)( x+h )+√ ( x+h)2−1−x−√ x2−1)

límh→0

1h ( (x+h )+√( x+h )2−1−x−√ x2−1

(x+√x2−1 ) )(1)

límh→ 0

1h ( h+√ ( x+h )2−1−√ x2−1

(x+√x2−1 ) )límh→0

1h ( h

( x+√x2−1 ) )+límh→0

1h (√ ( x+h )2−1−√x2−1

(x+√ x2−1 ) )multiplicamos por suconjugada

B'(x )=( 1

(x+√ x2−1 ) )+ límh→0

1h

(√( x+h )2−1−√ x2−1)(x+√ x2−1 )

(√( x+h )2−1+√x2−1)(√( x+h )2−1+√x2−1)

B'(x )=( 1

(x+√ x2−1 ) )+ límh→0

1h

( ( x+h )2−1−x2+1 )(x+√x2−1 ) (√( x+h )2−1+√x2−1)

B'(x )=( 1

(x+√ x2−1 ) )+ límh→0

1h

(x2+2xh+h2−1−x2+1 )(x+√x2−1 ) (√( x+h )2−1+√x2−1)

B'(x )=( 1

(x+√ x2−1 ) )+ límh→0

1h

(2xh+h2 )(x+√x2−1 ) (√( x+h )2−1+√x2−1)

25



B'(x )=( 1

(x+√ x2−1 ) )+ límh→0

1h

h (2 x+h )

(x+√x2−1 ) (√( x+h )2−1+√x2−1)

B'(x )=( 1

(x+√ x2−1 ) )+ límh→ 0

(2x+h )

(x+√ x2−1 ) (√ ( x+h )2−1+√x2−1)

remplazamos h→0

B'(x )=( 1

(x+√ x2−1 ) )+ límh→ 0

(2x+0 )

(x+√ x2−1 ) (√ ( x+0 )2−1+√x2−1)

B'(x )=( 1

(x+√ x2−1 ) )+ 2x

(x+√ x2−1 ) (√ x2−1+√x2−1 )

B'(x )=( 1

(x+√ x2−1 ) )+ 2 x

(x+√ x2−1 )2√x2−1B'

(x )=1

( x+√x2−1 )+ x

(x+√ x2−1)√x2−1

factorizamos

B'(x )=

1

( x+√x2−1 ) (1+x

√ x2−1 )B'

(x )=1

( x+√x2−1 ) ( x+√x2−1√ x2−1 )

B'(x )=

1

√x2−1∴ arccosh ( x )= 1

√ x2−1

U) .......................... (Joseph Michael Cayllahua Ch)

y=arctanh ( x )

x=tanh ( y )

x= e y−e− y

e y+e− y

26


ESCUELA ACADÉMICO PROFESIONAL DE INGENIERÍA CIVIL - HUANCAVELICAdespejando y :

y=12ln( x+11−x )

y '=limh→0

[ 12 ln( x+h+11−( x+h ) )−12 ln( x+11−x )]

h

y'=limh→0

1

2hln(

x+h+11−( x+h )x+11−x

)y '=

limh→0

1

2hln( ( x+h+1 ) (1−x )

(1−( x+h ) ) ( x+1 ) )

y '=limh→0

1

2hln(−x2+h−hx+1

−x2−h−hx+1 )

y '=limh→0

1

2hln(1+ 2h

−x2−h−hx+1 )

α= 2h

−x2−h−hx+1

y '=limh→0

1

2hln (1+α )α 1

α

y '=limh→0

α

2hlimh→0

ln (1+α )1α

27



Por limites notables :

y '=limh→ 0

2h

2h (−x2−h−hx+1 )

y '=limh→0

1

−x2−h−hx+1

C ' ( x )= 1

1−x2

II. DIFERENCIALES:

1. Haciendo uso de la diferencial hallar lo siguiente:

a) 3√25…………………………………(Joe Johan Zúñiga Curipaco)

f ( x )= 3√x

Sea:x0=27

f (x0+h )=f (x0 )+ f ' (x0 )h

f (27−2 )=f (27 )+ f ' (27 ) (−2 )

3√25= 3√27− 2

33√(27 )2

3√25≅ 3− 227

3√25≅ 2.9259

b) 3√67 …………………………………(Joe Johan Zúñiga Curipaco)

f ( x )= 3√x

Sea:x0=64

f (x0+h )=f (x0 )+ f ' (x0 )h

28



f (64+3 )=f (64 )+ f ' (64 ) (3 )

3√67=3√64+ 13√(64 )2

3√67≅ 4+ 116

3√67≅ 4.0625

c) 3√83d) 4√17

e) 4√15 …………………………………(Eliseo Silvestre de la Cruz ) f ( x+∆x )=f (16−1)

f ( x )= 4√x=4√16

( x )= 1

44√x3

= 14√(16)3

= 132

h=−1

f ( x+∆x )=f ( x )+ f ' ( x )∗h

f (16+1 )= f (16 )+ f ' (16 )∗(−1)

4√15=4√16+ 14√163

∗(−1)

3√15≅ 2− 132

3√15≅ 6332

≅ 1.969

f) sen(28 °)…………………………………(Eliseo Silvestre de la Cruz )

Solución:

29


ESCUELA ACADÉMICO PROFESIONAL DE INGENIERÍA CIVIL - HUANCAVELICAf ( x )=sen ( x ) , x0=30° ;h=−2°

f '( x)=cos ( x )

f (x0+h )≅ f (x0 )+hf ' (x)

f (30 °−2° )≅ sen (30 ° )+cos (30 ° ) (−2 ° )

sen (28 ° )≅ 12−( 2π180 )(√32 )

sen(28 °)≅ 0.47

g) cos (27 ° )h) tan (32 ° )i) cot (29 ° )j) sec (62 ° )k) ln (0.00000005 )

l) e0.9

f ( x+h )−f ( x )=h . f ´ ( x )

En donde

f ( x )=ex

Y se halla

f ´ ( x )=ex

e (1−0.1 )=e+ (−0.1 ) e1

e0.9=2.718281-0.271828

e0.9=2.446453828

30



m) arctan (0.97 )

Solución:

f ( x )=arctan ( x ) , x0=1 ;h=−0.03

f '( x)= 1

x2+1

f (x0+h )≅ f (x0 )+hf ' (x)

f (1−0.03 )≅ arctan (1 )−0.03( 1

x2+1 )

arctan (0.97 )≅ 45−0.032

arctan (0.97 )≅ 44.985

2. Hallar la derivada implícita de las funciones :

a) f ( x , y )=3√ x y2+cos3 ( xy )+ex sen (xy ) …………………………………(Joe Johan Zúñiga

Curipaco)

En función de “x”

∂ f∂ x

= ∂∂ x

( 3√ x y2 )+ ∂∂ x

(cos3 ( xy ) )+ ∂∂x

(ex sen ( xy) )

∂ f∂ x

= ∂∂ x

(x y2 ) 1

33√(x y2 )2

+3cos2 ( xy ) ∂∂x

(cos ( xy ) )+ ∂∂ x

(x sen ( xy ) ) (ex sen ( xy ))

∂ f∂ x

= y2

33√( x y2 )2

−3cos2 ( xy ) sen ( xy ) y+( sen ( xy )+xcos (xy ) y ) (e x sen( xy ) )

∂ f∂ x

= y2

33√( x y2 )2

−3 y cos2 ( xy ) sen ( xy )+( sen ( xy )+xcos (xy ) y ) (e x sen( xy ) )

31


ESCUELA ACADÉMICO PROFESIONAL DE INGENIERÍA CIVIL - HUANCAVELICAEn función de “y”

∂ f∂ y

= ∂∂ y

( 3√x y2 )+ ∂∂ y

(cos3 (xy ))+ ∂∂ y

(ex sen ( xy ) )

∂ f∂ y

= ∂∂ y

(x y2 ) 1

33√ (x y2 )2

+3cos2 ( xy ) ∂∂ y

(cos ( xy ) )+ ∂∂ y

(x sen ( xy ) ) (ex sen ( xy ))

∂ f∂ y

= 2 xy

33√ (x y2 )2

−3cos2 ( xy ) sen (xy ) x+(x cos ( xy ) x ) (e x sen( xy ) )

∂ f∂ y

= 2 xy

33√ (x y2 )2

−3 x cos2 ( xy ) sen ( xy )+(x2 cos ( xy ) ) (ex sen ( xy ))

f ' ( x , y )=

−∂ f ( x , y )∂ x

∂ f (x , y )∂ y

¿

− y2

33√ (x y2 )2

+3 ycos2 (xy ) sen ( xy )−( sen ( xy )+x cos ( xy ) y ) (ex sen ( xy ) )

2xy

33√(x y2 )2

−3 x cos2 ( xy ) sen ( xy )+(x2 cos ( xy ) ) (e x sen( xy ) )

b) g ( x , y )=xy−2+3√cot4 (x2 y )−x+ex

3 tan ( xy )

c) h ( x , y )=ln (3 xe xy )+ 3√sech2 (x2 y−1 )+3 x

d) i (x , y )=( 4√ y2 x−3 )3+3√arccos3( x y 13 )−x2 y .................... (Joseph M. Cayllahua)

Mediante:

f ' ( x , y )=

−∂∂ x

f ( x , y )

∂∂ y

f ( x , y )

32


ESCUELA ACADÉMICO PROFESIONAL DE INGENIERÍA CIVIL - HUANCAVELICARESOLUCIÓN

En:

∂∂ x

[ ( 4√ y2 x−3 )3+3√arccos3(x y 13)−x2 y ]

∂∂ x

( 4√ y2 x−3 )3+ ∂∂ x

3√arccos3( x y 13 )−x2 y

3

4 ( 4√ y2 x−3 )∂∂ x

( y2 x−3 )+ 1

3 [ 3√(arccos3(x y 13)−x2 y)2]

∂∂ x

[arccos3( x y 13 )−x2 y ]

−94

4√ y6

x13−

3√arccos3(x y 13)−x2 y

3[arccos3(x y 13 )−x2 y ] [ 3 ( 3√ y )arccos2(x y13)√1−(x y

13)2

1−(x y 13)2 +2 xy ]

En:

∂∂ y

[ ( 4√ y2 x−3 )3+3√arccos3( x y 13 )−x2 y ]

∂∂ y

( 4√ y2 x−3 )3+ ∂∂ y

3√arccos3 (x y 13)−x2 y

3

4 ( 4√ y2 x−3 )∂∂ y

( y2 x−3 )+ 1

3[ 3√(arccos3( x y 13 )−x2 y)2]

∂∂ y

[arccos3(x y 13)−x2 y ]

324√ y2

x9−


3[arccos3( x y 13 )−x2 y ] {x ( 3√ y )arccos2(x y13)√1−( x y

13 )2

y [1−( x y13 )2]

+x2}∴

33



i' ( x , y )=

−{−94 4√ y6

x13−


3[arccos3( x y 13 )−x2 y ] [ 3 ( 3√ y )arccos2(x y13)√1−( x y

13 )2

1−(x y13)2

+2 xy ]}324√ y2

x9−


3 [arccos3( x y 13 )−x2 y ] {x ( 3√ y )arccos2( x y

13 )√1−( x y

13 )2

y [1−(x y13)2]

+ x2}RPTA:

i' ( x , y )=

944√ y6

x13+


3 [arccos3 (x y13 )−x2 y ] [ 3 ( 3√ y )arccos2( x y13 )√1−( x y

13 )2

1−( x y 13 )2 +2xy ]

324√ y2

x9−

3√arccos3(x y 13)−x2 y

3 [arccos3(x y 13)−x2 y ] {x ( 3√ y )arccos2 (x y13 )√1−(x y

13)2

y [1−( x y13 )2]

+x2}e) f ( x , y )=senh2 (xy )+arcsec ( 3√sech2 ( x√ y3 ))

f) k ( x , y )=arccos (x y3 )+arccosh (tan (x y3) ) (henry)

SOLUCION:

d (k ( x ; y ) )=−( ∂ (arccos (x y3 )+arccosh ( tan(x y3) ))

∂ x )( ∂ (arccos (x y3 )+arccosh ( tan(x y3) ) )

∂ y )

d (k ( x ; y ) )=−( ∂ (arccos (x y3 ))

∂ x+∂ (arccosh ( tan(x y3) ) )

∂ x )( ∂ (arccos (x y3 ))

∂ y+∂ (arccosh ( tan(x y3) ) )

∂ y )

34



d (k ( x ; y ) )=

−( − y3

√1−( x y3)2+

± ( y3 sec2(x y3))

√( 3√sech2(x√ y3))2+1 )( −3 xy2

√1−( x y3)2+

± (3xy2 sec 2(x y3))

√( 3√sech2(x√ y3))2+1 )(+si arccosh (x y3 )>0 ; tan (x y3 )>1−si arccosh ( x y3 )<0 ; tan (x y3 )>1)

d (k ( x ; y ) )=

−(− y3√( 3√sech2(x√ y3))2+1+√1−(x y3)2 (± ( y3 sec 2(x y3)) )

√( 3√ sech2(x √ y3))2

+1 (√1−(x y3)2 ) )(−3 xy2√ ( 3√sech2(x√ y3))2+1+√1−(x y3)2 (± (3 xy2 sec2(x y3)))

(√ ( 3√sech2(x √ y3))2

+1)√1−(x y3)2 )

d (k ( x ; y ) )=(− y3√( 3√sech2( x√ y3))2+1+√1−(x y3)2 (± ( y3 sec2(x y3)) ))

(3 xy2√ ( 3√sech2(x √ y3))2+1−√1−(x y3)2 (± (3 xy2 sec2(x y3))) )

III. MÁXIMOS Y MÍNIMOS

En las funciones siguientes hallar:

a) Intervalos de concavidadb) Máximos y mínimos relativosc) Máximos y mínimos absolutosd) Puntos de intersección con el eje Xe) Puntos de inflexión

2¿ . g ( x )=(x2−9)(x2−81)

SOLUCION:

PrimeraDerivada :

g ' ( x )=2 x (x2−81 )+2x (x2−9)

35


ESCUELA ACADÉMICO PROFESIONAL DE INGENIERÍA CIVIL - HUANCAVELICA g ' ( x )=2 x3−162x+2 x3−18 x

g ' ( x )=4 x3−180x=0

x (4 x2−180)=0

x1=0 ; x2=3√5 ; x3=−3 √5

Segunda Derivada :

g ' ( x )=4 x3−180x

g ' ' ( x )=12 x2−180

⇝ g' ' (0 )=12 (02 )−180

g' ' (0)=−180;−180<0maximorelativo

g (0 )=(02−9)(02−81)

g (0 )=729

∴(0 ;729)

⇝ g' '¿¿

g' '¿ ¿

g (3√5 )=[ (3√5 )2−9 ] [ (3√5 )2−81 ]

g (3√5 )=(36)(−36)

g (3√5 )=−1296

∴(3√5 ;−1296)

⇝ g' '¿¿

g' '¿ ¿

g (−3√5 )=[ (−3√5 )2−9 ] [ (−3√5 )2−81 ]

g (−3√5 )=(36)(−36)

36


ESCUELA ACADÉMICO PROFESIONAL DE INGENIERÍA CIVIL - HUANCAVELICA g (−3√5 )=−1296

∴(−3√5;−1296)

Punitos de Inflexion :

g ' '(x)=12 x2−180=0

12 x2−180=0

x2=15

x=±√15

x4=√15∧ x5=−√15

⇝ g ( x )≅ (x2−9)(x2−81)

g (√15 )≅ (√152−9)(√152−81)

g (√15 )≅−396

g (−√15 )≅ [ (−√15 )2−9 ] [ (−√15 )2−81 ]

g (−√15 )≅−396

∴los puntos de inflexion son :

(√15 ;−396)∧(−√15;−396)

Concavidad :

( f ¿¿ ' ' ( x )>0 ; f ' ' ( x )<0)¿

12 x2−180>0

(x+√15 ) (x−√15 )>0

(x>√15)∧¿

37


ESCUELA ACADÉMICO PROFESIONAL DE INGENIERÍA CIVIL - HUANCAVELICA+ - +

−√15 √15

f esconcava haciaarriba :

←∞ ;−√15>u<√15 ;∞>¿

f esconcava haciaarriba :

←√15; √15>¿

→x4−90 x2+729=0

x2=m

m2−90m+729=0

m=90±√902−4(729)

2

m=90±722

m1=81∧m2=9

x2=m

x=±9∧ x=±3

Grafica :

38



IV.IV.IV.IV.IV.IV.IV.IV.IV.IV.IV.IV.IV.IV.IV.IV.IV.IV.IV.IV.IV.IV.IV.IV.IV.IV.IV.IV.IV.IV.IV.IV.IV.IV.IV.

METODOS DE INTEGRACION INDEFINIDA 1. METÓDO DE DESCOMPOSICION

a¿∫ dx

2x2+5 x+2…………………………………(Joe Johan Zúñiga Curipaco)

(0,729)

-9 -3 3 9

(−√15 ;−396) (√15 ;−396)

(−3√5 ;−1296) (3√5 ;−1296)

39


ESCUELA ACADÉMICO PROFESIONAL DE INGENIERÍA CIVIL - HUANCAVELICA13∫

[ (2x+4 )− (2 x+1 ) ] dx(2x+1 ) ( x+2 )

13∫

[ (2 x+4 ) ]dx(2x+1 ) ( x+2 )

−13∫

[ (2x+1 ) ]dx(2 x+1 ) ( x+2 )

13∫

2 ( x+2 )dx(2x+1 ) ( x+2 )

−13∫

dx( x+2 )

13∫

2dx(2x+1 )

−13∫

dx(x+2 )

13ln|2 x+1|+C1−

13

( ln|x+2|+C2 )

13ln|2 x+1x+2 |+K

b¿∫ dx( x+3 ) (4+3 x ) …………………………………(Joe Johan Zúñiga Curipaco)

15∫

[ (3x+9 )−(4+3 x ) ]dx( x+3 ) (4+3 x )

15∫

[ (3 x+9 ) ] dx(x+3 ) (4+3 x )

−15∫

[ (4+3 x ) ] dx( x+3 ) (4+3 x )

15∫

3 ( x+3 )dx(x+3 ) (4+3 x )

−15∫

dx( x+3 )

15∫

3dx(4+3x )

−15∫

dx( x+3 )

15ln|4+3 x|+C1−

15

¿

40



15ln|4+3 xx+3 |+K

d ¿∫ x3dx2 x2−3 x+5

…………………………………(Eliseo Silvestre de la Cruz)

∫( 12 x+ 34− x

4 (2x2−3 x+5 )−

15

4 (2x2−3 x+5 ) )dx12∫ xdx+¿ 3

4∫ dx−¿ 14∫

x

(2 x2−3 x+5 )dx−15

4 ∫ dx

(2 x2−3 x+5 )¿¿

14x2+ 3

4x− 14∫

x

(2 x2−3x+5 )dx−15

4 ∫ dx

(2 x2−3x+5 )

¿14x2+

34x−

12 i√31 {∫ x

x−( 3+i√314 )dx−∫ x

x−( 3−i √314 )

dx }− 152i √31 {∫ 1

x−( 3+i √314 )dx−∫ 1

x−( 3−i √314 )

dx}¿ 14x2+ 3

4x− 1

2 i√31¿

41



¿ 14x2+ 34x−( 3+i√318 i√31 ) ln|x−( 3+ i√314 )|−( 3−i √31

8 i√31 ) ln|x−( 3−i√314 )|−3+i√318 i√31

ln|x−(3+ i√314 )|−3−i√318 i√31

ln|x−( 3−i√314 )|…Rpta

e ¿ .∫ x2dx(x+3)(x−4)(x+5)

…………………………………(Eliseo Silvestre de la Cruz)

SOLUCIÓN:

∫ (x2−x−12 )dx(x2−x−12)(x+5)

+¿∫ ( x+12 )dx(x2−x−12)(x+5)

¿

∫ dx(x+5)

+¿∫ ( x+5+7 )dx( x2−x−12)(x+5)

¿

∫ dx(x+5)

+¿∫ ( x+5 )dx( x2−x−12)(x+5)

+7∫ dx

( x2−x−12)(x+5)¿

42



∫ dx(x+5)

+¿∫ dx

( x2−x−12)+79∫

( x+5−x+4 )dx(x2−x−12)(x+5)

¿

∫ dx(x+5)

+¿∫ dx

( x2−x−12)+79∫

( x+5 )dx(x2−x−12)(x+5)

−79∫

( x−4 )dx(x2−x−12)(x+5)

¿

∫ dx(x+5)

+¿∫ dx

( x2−x−12)+79∫

dx

(x2−x−12)−79∫

dx(x+3)(x+5)

¿

∫ d (x+5)(x+5)

+¿∫ dx

(x−12 )2

−14−12

+79∫ dx

(x−12 )2

−14−12

−79∫ dx

( x+4 )2−16+15¿

∫ d (x+5)(x+5)

+¿∫ dx

(x−12 )2

−( 72 )2+79∫ dx

(x−12 )2

−( 72 )2−79∫ dx

( x+4 )2−1¿

ln (x+5)+c1+17ln|x−12−72x−

12+72

|+c2+ 17 . 79 ln|x−12−72x−12+72

|+c3−12 . 79 ln|x+4−1x+4+1|+c4

c1+c2+c3+c4=K

ln (x+5)+ 17ln|x−4x+3 |+ 19 ln|x−4x+3 |− 7

18ln|x+3x+5|+K

f) ...........

g) ∫ dx

[ sen2 (x )−1 ] [ cosec2 ( x )−1 ] .......................... (Joseph M. Cayllahua Ch.)

RESOLUCIÓNDATOS:

1+cot2 ( x )=cosec 2 ( x )

43



sen2 ( x ) +cos2 ( x )=1

∫ dx

[ sen2 (x )−1 ] [ cosec2 ( x )−1 ]=∫ −dx

[1−sen2 ( x ) ] [ cot2 ( x ) ]

−∫ tan2 ( x )dxcos2 ( x )

=−∫ tan2 ( x ) sec2 ( x )dx

−∫ tan2 ( x )d [ tan ( x ) ]=−¿tan3 ( x )3

+k ¿

∴

RPTA: −tan3 ( x )3

+k

2. METÓDO DE INTEGRACION POR PARTES

c ¿ .∫ x5 ex3

dx

Solución:

∫ x3 x2 ex3

dx

Integramos por partes :

u=x3 dv=x2 ex3

dx

du=3 x2dx ∫ dv=∫ x2e x3dx

v=( x3 )e x3

3 (x3 )

v= ex3

3

∫udv=u . v−∫ vdu

∫ x5 ex3

dx= x3 ex3

3−∫ e x33 x2dx

3

∫ x5 ex3

dx= x3 ex3

3−∫ ex

3

x2dx

∫ x5 ex3

dx= x3 ex3

3− ex

3

3+c

44



g¿∫ ln( ln( 2x3 ))d (x)

u=ln( ln 2 x3 ) dv=x

du=dx3x ¿¿ v=dx

x ln( ln 2 x3 )-3∫ x2 ln( 2 x3 )dx u=ln( 2 x3 ) dv=x2dx

u=dx3 x

v= x3

3

x ln( ln 2 x3 )-3( x33

ln( 2 x3 ))-13∫ xx

3

dx

x ln( ln 2 x3 )-(x3 ln( 2 x3 ))-13∫ x2dx

x ln( ln 2 x3 )-(x3 ln( 2 x3 ))-13 x33

+k

h) ∫ ln (sen( √ x2

))dx (henry)

Solución:

∫ ln (sen( √ x2 ))dx=∫ ln (sen( x12

2))dx

∫2uln (sen( u2))du ;dondeu=x

12

2∫ 4 vln (sen (v))dv ;donde v=u2

8∫vln (sen (v ))dv=8(lnsen (v ) 12v2−∫ 12 v

2 ddvln ( sen ( v ) )dv)

8( 12 v2lnsen (v )−12∫ v2

ddvln (sen (v ) )dv )=4 v2 lnsen ( v )−4∫ v2

ddvln (sen (v ) )dv

4 v2 lnsen ( v )−4∫ v2 cotvdv=4 v2lnsen (v )−4 (v2 ln|senv|−∫ ln|senv| ddv v2dv)

45



4 (v¿¿2lnsen (v )+∫ ln|senv| ddv v2dv−v2 lnsen ( v ))¿=

4 v2 lnsen ( v )+4∫ ln|senv| ddv v2dv−4 v2lnsen (v ) ¿

4 v2 lnsen ( v )+4∫ ln|senv|2vdv−4 v2lnsen (v )¿=4 v2lnsen (v )+8∫ vln|senv|dv−4 v2lnsen (v ) ¿

4 v2 lnsen ( v )+8( ln|senv|12v2−∫ 12 v

2 ddvln (sen (v ) )dv ¿−4v2 lnsen ( v )¿)

4 ¿¿

4 v2 lnsen ( v )−4∫ v2cosvsenv

dv=4 v2lnsen (v )−4∫ v2 cosvsenv

dv

i). ∫ x3dx

(x2−2)3 (henry)

Solución:

∫ x3dx(x2−2)3

=∫ u+22u3

du ;donde u=¿ x2−2¿

12∫

u+2u3

du=¿ 12∫( u

u3+ 2u3

)du=¿ 12∫

1

u2du+∫ 1

u3du¿¿

12 (−1u )− 1

2u2+c1=−1

2u− 1

2u2+c 1

Remplazando u=x2−2−1

2 (x2−2 )− 1

2 (x2−2 )2+c1= −1

2 ( x−√2 )(x+√2)− 1

2 (x2−2 )2+c1

−12 (x−√2 ) (x+√2 )

− 1

2 ((x−√2 ) (x+√2 ) )2+c1

−12 (x−√2 ) (x+√2)

− 1

2 ( x−√2 )2 (x+√2 )2+c1

−12 (x−√2 ) (x+√2)

− 1

2(x2−2√2 x+2) (x+√2 )2+c1

−12 (x−√2 ) (x+√2)

− 1

2(x+−2√22

+ √02

)(x+−2√22

−√02

)(x+√2 )2+c1

−12 (x−√2 ) (x+√2)

− 1

2 ( x+√2 )2(x+−2√22 )

2+c1

−12 (x−√2 ) (x+√2)

− 12¿¿¿

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ESCUELA ACADÉMICO PROFESIONAL DE INGENIERÍA CIVIL - HUANCAVELICA−1

2 (x−√2 ) (x+√2)− 1

2 ( x+√2 )2(x¿¿2+2√2 x+2)+c 1¿

−12 (x−√2 ) (x+√2)

− 1

2(x+−2√22

+ √02

)(x+−2√22

−√02

)(x¿¿2+2√2 x+2)+c 1¿

−12 (x−√2 ) (x+√2)

− 1

2(x+−2√22

)(x+−2√22

)(x¿¿2+2√2 x+2)+c1¿

−12 (x−√2 ) (x+√2)

− 1

2( x+−2√22 )

2

(x¿¿2+2√2 x+2)+c 1¿

−12 (x−√2 ) (x+√2)

− 1

2( x−2√22 )2

(x¿¿2+2√2 x+2)+c1¿

−12 (x−√2 ) (x+√2)

− 1

2( x−2√22 )2

(x+√2 )2+c1

−12 (x−√2 ) (x+√2)

− 1

2( x−2√22 )2

(x¿¿2+2√2 x+2)+c1¿

−12 (x−√2 ) (x+√2)

− 1

2 ( x−√2 )2(x¿¿2+2√2 x+2)+c1¿−1

2 (x−√2 ) (x+√2)− 12(x¿¿2−2√2 x+2)(x¿¿2+2√2x+2)+c1¿¿

−12 (x−√2 ) (x+√2)

− 1

2(x+−2√22

+ √02

)(x+−2√22

−√02

)(x¿¿2+2√2 x+2)+c 1¿

−12 (x−√2 ) (x+√2)

− 1

2(x+−2√22

)(x+−2√22

)(x¿¿2+2√2 x+2)+c1¿

−12 (x−√2 ) (x+√2)

− 1

2( x+−2√22 )

2

(x¿¿2+2√2 x+2)+c 1¿

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ESCUELA ACADÉMICO PROFESIONAL DE INGENIERÍA CIVIL - HUANCAVELICA−1

2 (x−√2 ) (x+√2)− 1

2( x+−2√22 )

2

(x+√2 )2+c1

O.-∫ ln( x)x3

dx..................... (Joseph M. Cayllahua Ch.)

RESOLUCIÓNsea :

u=ln ( x ) dv=1

x3dx

du=1xdx v=∫ 1

x3dx

v=−12x2

∴−ln ( x )2 x2

−∫(−12x2 )( 1x )dx−ln ( x )2 x2

+∫ 1

2x3dx

−ln ( x )2 x2

− 1

4 x2+k

RPTA: −[2 ln (x )+1 ]

4 x2+k

48



3. METÓDO DE INTEGRACIÓN TRIGONOMETRICA

a) ∫cos3(2 x)sen (3 x)dx

Solución:

∫(cos2 x−sen2 x)3 sen (3 x)dx ; Donde: cos2 x=cos2 x−sen2 x

∫ (cos2 x−sen2 x )3 (3 senx−4 sen3 x )dx ;

Donde: sen3 x=3 senx−4 sen3 x

∫¿¿

−∫(4u2−1)(2u2−1)3du

−∫ (4u2−1 ) (2u2 )3−3 (2u2 )2+6u2−1¿¿du

−∫ (4u2−1 )(8u6−12u4+6u2−1)du

−∫(32u8−56u6+36u4−10u2+1)du

−32∫ u8du+56∫u6du−36∫ u4du+10∫u2du−1∫du

−32 u9

9+56 u

7

7−36 u

5

5+10 u

3

3−u+c1

−329

u9+ 567u7−36

5u5+ 10

3u3−u+c1 ;reelplazandou=cosx

−329cos9 x+8cos7 x−36

5cos5 x+10

3cos3 x−cosx+c1

49



b) ∫cos4 (2x ) tan2(x)dx

∫cos4 (2x )¿¿

∫ sen2(x )cos2(x )dx

∫ sen2(x )cos (2 x )+1

2dx Aplicando que: cos2(x)=

cos (2 x )+12

∫ 1−cos (2x )2

cos (2 x )+12

dx Aplicando que: sen2(x )= 1−cos (2 x )

2

∫¿¿¿ Dónde: u=2x

18∫(1−cos (u))(1+cos (u))dx

18∫(1−cos (u))(cos (u )+1)dx

18∫(1−cos2(u))dx

18(∫ 1du−∫cos2 (u )du)

18∫1du−

18∫ cos

2 (u )du

18u−18∫

cos (2u )+12

du

18u− 116∫cos (2u )du− 1

16∫ 1du

116

u+−132

sen (2u )+c1 De donde reemplazando u=2x tenemos:

18x− 132

sen (4 x )+c1

50



c) ∫cot3(2 x)cos2(2 x)dx

Solución:

∫ cos2(u)cot3(u)2

du ;dondeu=2x

12∫ cos

2(u)cot3(u)du

12∫ cos2(u) cos

3usen3u

dusabiendoque : cot3 (u )= cos3u

sen3u

12∫ cos5u

sen3udu

12∫ ¿¿

−12∫ v5u

(1−v¿¿2)2dv¿

−12∫( −w5

2(w+1)2 )dw ;dondew=−v2

12∫ w5

2 (w+1)2dw

14∫ (s−1)2

s2ds ;donde s=w+1

14∫ s2−2 s+1

s2ds

14∫( s

2

s2−2 ss2

+ 1s2

)ds

14∫ 1ds−

12∫

1sds+ 1

4∫1

s2ds

51



14s−14ln ( s )− 1

4 s+c1

Reemplazando: s=w+1

14(w+1)−1

4ln (w+1 )− 1

4 (w+1)+c1

14w−1

2ln (w+1 )+ 1

4− 14 (w+1)

+c 1

14w−1

2ln (w+1 )− 1

4 (w+1 )+c2

Reemplazando: w=−v2

14(−v2)−1

2ln (−v2+1 )− 1

4 (−v2+1 )+c2

−14v2−1

2ln (1−v2 )− 1

4 (1−v2)+c 2

−14v2−1

2ln (1−v2 )− 1

4 (1−v )(1+v )+c 2

Reemplazando: v=cosu−14cos2u−1

2ln (1−cos2u )− 1

4 (1−cosu )(1+cosu )+c 2

−14cos2u−1

2ln (sen2u )− 1

4 (1−cosu )(1+cosu)+c2

−14cos2u−1

22 ln ( senu )− 1

4 (1−cosu)(1+cosu)+c2

−14cos2u−ln (senu )− 1

4 (1−cosu )(cosu+1)+c 2

Reemplazando: u=2x−14cos22 x−ln ( sen2x )− 1

4 (1−cos 2x )(cos 2x+1)+k

52



d) ∫ cosec3(2x ) tan3(2 x)dx (henry)

Solución:

∫ tan3(u)cosec 3(u)2

du ;dondeu=2x

12∫ tan

3(u)cosec3(u)du

12∫ ¿¿

12∫

sen3(u) 1

sen3(u)cos3(u)

du

12∫

1

cos3(u)du

12∫ ses3(u)du

12¿

14sec (u ) tan (u )+ 1

4∫ sec(u)du

14sec (u ) tan (u )+ 1

4ln (sec (u )+ tan (u ) )+c 1

Reemplazando: u=2x

14sec (2 x ) tan (2x )+ 1

4ln ( sec (2 x )+ tan (2 x ) )+c1

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Primera Práctica de Análisis Matemático II

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Transcript of Primera Práctica de Análisis Matemático II