( 32 x )= 53 arcsen ( 32 x ) sen x +7 cos x 1 dx. x x. e 2 x +1 dx. 5x 7 dx. x sen x dx. x 4 x x 1 dx. x 2 dx. dxx. x x x dx. 1 x.

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(1)

I.E.S. Juan Carlos I

Ciempozuelos (Madrid)

Matemáticas II

*

Análisis III: Integrales

*

1. Integrales inmediatas (o casi inmediatas):

a)

4x

2

– 5x

7

dx

=4· 1 3x 3– 5· 1 2x 2+7 x +C= 4 x3 3 – 5 x 2 2 +7 x + C

b)

3 x

3

– 5 x

2

+

3 x +4 dx

=3· 1 4x 4– 5· 1 3x 3+3· 1 2x 2+4 x +C= 3 x4 4 – 5 x 3 3 + 3 x2 2 +4 x + C

c)

1

2 x

7

dx

= 1 2ln∣2 x + 7∣+ C =ln

2 x +7∣+C

d)

5

dx

x

x −15 dx = 5 4x 4 5+C= 5 5

x4 4 +C

e)

x – sen x

dx

x dx+

sen x dx=1 2x 2 +cos x +C

f)

sen x+7 cos x – 1 dx

sen x dx+7

cos x dx−

dx=−cos x +7 sen x − x +C

g)

tg

2

x dx

1+tg2x −1dx=

1+tg2x dx−

dx=tg x − x +C

h)

x – 2 dx

x 1 2dx−2

dx=2 3x 3 2−2x= 2

x3 32 x +C

i)

2

x

dx

4 2

xdx= 4

1 2

xdx=4

x +C

j)

e

2 x +1

dx

122e2 x + 1dx=1 2

2e 2 x +1dx=1 2e 2 x +1 +C

k)

cos(a x +b)dx

1aa· cos(a x +b)dx=1

a

a· cos(a x +b)dx= 1 asen(a x+ b)+C

l)

x

2

4 x



x

2

1

dx

=

x4+4 x3 −x2−4 x dx=1 5x 5 +x41 3x 32 x2 +C

m)

3

x

2

dx

= 1 3

2

x 1 3dx= 1 3

2 3 4x 4 3+C=3 4 3

x4 2 +C

n)

x−1

dx

=ln∣x− 1∣+ C

o)

x

x

x

2

dx

=

1x+

xx2dx=

1 xdx +

x −3 2dx=ln∣x∣− 2 x1 2+C

p)

3

1

x

2

dx

=3

1 1+x2dx=3arctg x +C

q)

5

2 – 3 x

2

dx

= 5

2

1

1−3 2x 2 dx= 5

2

1

1−

(

3 2x

)

2dx= 5

2

2 3arcsen

(

3 2x

)

= 5

3arcsen

(

3 2x

)

r)

1

(

x−2)

2

dx

=

(x−2) −2dx=−( x−2)1+C=− 1 x −2+C

(2)

s)

e

2x −3

dx

=1 2e 2 x −3 +C

t)

5 dx

1+7 x

2 =5

1 1+(

7x )2dx= 5

7arctan(

7x )+C

u)

x

2

– 5 x+4

x +1

dx

x2 = −5 x+ 4=(x−6 )(x +1)+ 10

x−6+ 10 x+1dx= 1 2x 26x +10 ln∣x+1∣+C

v)

x

3

– 3x

2

x – 1

x

2

dx

x3 = −3 x2 +x+1=(x−2)(x2 +x+3)+5

x 2 +x+3+ 5 x −2dx= x 3 3+ x 2 2+3x +5ln∣x− 2∣+C

w)

x – 1

2 x –

x +1

dx

=

(x –1)(

2x +

x +1) (

2x –

x +1)(

2 x+

x+1)dx =

(x –1)(

2x +

x +1) x−1 dx=

(

2x +

x +1)dx= 2

2 3 x 3 2+2 3(x+ 1) 3 2+C

x)

e

x

(

x−1) – e

x

(

x−1)

2

dx

Derivada de un cociente= ex x −1+C

y)

2 e

2 x

+e

2 x

e

x

dx

=

3 exdx =3ex+C

z)

3

11

x

dx

=3

e −ln 11· xdx =− 3 ln11eln11 · x+C = 3 11xln11+C

2. Integrales de funciones compuestas / cambio de variable:

a)

(2 x+6)

5

dx

= t=2 x + 6 d t =2dx

t5dt 2= 12

t 5dt= 1 12t 6 +C=(2x +6)6 12 +C

b)

5

7 x – 9

dx

t=7 x −9= d t=7 dx

5 t dt 7= 57

1 t dt= 57ln∣t∣+C= 57ln∣7x −9∣+C

c)

3 x e

5 x2−7

dx

= t =5 x2−7 d t =10 x dx

103 etdt = 3 10e t+C=3 10e 5 x27+C

d)

x

(

5x+3)(5x−3)

dx

x 25 x2−9dx t =25 x=2−9 d t =50 x dx

501 ·1 tdt = 1 50ln∣t∣+C= 1 50ln

25 x 29

+C

e)

cos x sen

3

x dx

= t=sen x d t =cos x dx

t3dt = 1 4t 4+C= sen4x 4 +C

f)

x

3

x

4

– 1

dx

t2==x4+1 2 t d t= 4 x3 dx

2tt dt=1 2

dt= 1 2t +C=

x41 2 +C

g)

sen x cos x dx

= t=sen x d t =cos x dx

t dt = 1 2t 2+C= sen2x 2 +C

h)

4 x – 3

2 x

2

– 3 x−14

dx

t =2 x2=−3 x −14 d t =(4 x−3)dx

1tdt =ln∣t∣+C=ln

2 x23 x −14

+C

i)

1

x ln x

dx

= t =lnx d t =dx x

1tdt=ln∣t∣+C=ln

(

ln∣x∣

)

+C

(3)

j)

x

(

x

2

+

3)

5

dx

t =x=2 +3 d t =2 x dx

t15 dt 2= 1 2

t −5dt =−1 8t −4 +C=− 1 8( x2+3)4+C

k)

1

x

ln

3

x dx

t =lnx= d t =dx x

t3dt =1 4t 4+C=1 4ln 4x∣+C

l)

2 x dx

9−x

2

dx

t=9− x= 2 d t =−2xdx

−1

tdt =−2

t +C=−2

9− x2+C

m)

sen x dx

cos

5

x

d t=−senx dxt =cos x=

−1 t5dt =−

t −5dt =1 4t −4 +C= 1 4cos4x+C

n)

x−1

x

2

– 2 x dx

= t2 =x2−2 x 2 t d t =(2 x −2)dx t dt =(x−1)dx

t2dt= 1 3t 3+C=(x22 x ) 3 2 3 +C

o)

arcsen x

1 – x

2

dx

= t =arcsenx d t = dx1−x2

t dt= 1 2t 2 +C= arcsen2x 2 +C

p)

1

ln x

2

x

dx

= t=1 +ln x d t = dx x

t2dt= 1 3t 3+C=(1 +ln x )3 3 +C

q)

x

2

x

3

1

3 5

dx

= t =x3 −1 d t=3 x2dx

t 3 5dt 3= 524t 8 5+C=5( x31) 8 5 24 +C

r)

x

4

e

x53

dx

= t =x5+3 d t=5 x4dx

etdt 5= 1 5e t+C=ex 5 +3 5 +C

s)

x sen

x

2

dx

= t =x2 + π d t =2 x dx

sen t dt 2=− 12cos t +C=− cos ( x2+ π ) 2 +C

t)

x ln

x

2

3

x

2

3

dx

= t =ln(x2 +3 ) d t = 2 x x2+3dx

t dt 2= 14t 2+C=ln2(x2+3) 4 +C

u)

e

x

1

e

2x

dx

= t =ex d t =exdx

1 1+t2dt=arctan t +C =arctan(e x)+C

v)

1

tg

2

x

tg x

dx

d t =(1 +tgt =tg x=2x )dx

1tdt =ln∣t∣+C=ln∣tg x∣+ C

w)

tg x dx

sen xcosxdx t =cosx=

d t =−senx dx

−1 tdt =−ln∣t∣+C=−ln∣cos x∣+C=ln∣sec x∣+C

x)

3

x

1

9

x

dx

3x 1 +(3x)2dx t =3=x d t=ln 3· 3xdx

ln31 · 1 1+t2dt= 1ln3arctant +C= arctan(3x ) ln 3 +C

y)

x

3

sen

x

4

– 3 

dx

= t =x4−3 π d t= 4 x3 dx

sen t dt 4=− 14cos t +C=− cos( x43π ) 4 +C

z)

e

x

5

x

dx

= t =x d t = dx 2√x

52etdt = 2 5e t+C= 2ex 5 +C

(4)

3. Integrales por partes:

a)

2 x e

x

dx

= u=2 xu ' =2 v '=exv =−ex −2x e−x+

2exdx=−2x ex– 2ex+C=−2( x +1) ex+C

b)

x e

− x3

dx

u =x =→u '=1 v '=e− x3 v=−3 e− x3 −3x e− x 3+

3ex 3dx=−3 x ex 3–9ex 3+C=−3( x +3)ex 3+C

c)

x

2

e

2 x

dx

= u=x2 →u ' =2x v ' =e2 xv = 1 2e 2 x 1 2x 2e2x

x e2xdx = u= xu '=1 v '=e2 xv= 1 2e 2 x 1 2x 2e2x −1 2x e 2x +1 2

e 2 xdx= =1 2x 2e2x1 2x e 2x+1 4e 2x+C=2 x22 x +1 4 e 2 x+C

d)

[

x e

x

]

3

dx

=

x3e3 xdx = u =x3 →u '=3 x2 v ' =e3 xv =1 3e 3 x 1 3x 3e3 x

x2e3xdx = u =x2 →u ' =2 x v '=e3 xv =1 3e 3 x =1 3x 3e3 x1 3x 2e3 x+2 3

x e 3 xdx = u= xu ' =1 v '=e3 xv=1 3e 3 x 1 3x 3e3 x1 3x 2e3 x+2 9x e 3 x2 9

e 3 xdx = 1 3x 3e3 x− 1 3x 2e3 x+ 2 9x e 3 x− 2 27e 3 x+C =9 x39 x2+6 x− 2 27 e 3 x+C

e)

x ln x dx

u=lnx=→u ' =1x v ' =xv =1 2x 2 1 2x 2ln x− 1 2

x dx = 12x 2ln x − 1 4x 2= x2 4 (2ln∣x∣− 1)+C

f)

x

2

ln xdx

u =lnx= →u ' = 1x v '= x2 →v =1 3x 3 1 3x 3ln x − 1 3

x 2dx = 1 3x 3ln x− 1 9x 3= x3 9 (3 ln∣x∣−1)+C

g)

ln x

x

dx

u =lnx=→u '=1 x v ' =1 xv=ln x ln2x−

ln x x dx

I

I=ln2x −I →2 I =ln2x+C →I =ln2x∣

2 +C

h)

x

x +1 dx

u= x =→u ' =1 v '=x+ 1 →v =2 3(x +1) 3 2 2 3x ( x+1) 3 22 3

(x +1) 3 2dx =2 3x ( x +1) 3 2 4 15(x +1 ) 5 2+C

i)

2 x

2 3

x +1 dx

= u =2 x2 →u '=4 x v '=3 √x+ 1 →v = 34(x +1 ) 4 3 3 2x 2 (x +1) 4 3 −

3 x (x +1) 4 3dx = u=3 xu '=3 v '=(x +1 ) 4 3 →v= 37(x +1 ) 7 3 =3 2x 2(x+1)349 7x ( x+1) 7 3+9 7

(x +1) 7 3dx=3 2x 2(x +1)439 7x ( x +1) 7 3+27 70(x+ 1) 10 3 +C

j)

e

x

sen x dx

= u =ex u '=ex v ' =sen x →v =−cos xcos x ex+

excos x d x

u=exu '=ex v ' =cosx →v=sen x =−cos x ex+sen x ex

exsen x dx

I

I=−cos x ex+sen x exI→ 2I=(sen x−cos x)ex+C→I =1

2(sen x −cos x ) e x +C

k)

e

x

2

x+1

dx

= u =2x + 1u '=ln 2 ·2x+ 1 v '=ex v=ex ln 2·ex2x +1−ln 2

ex2x +1dx

II =ln2ex2x +1−ln2· I →I= ln2 1+ln2e x2x +1

l)

3 x cos x dx

u=3 x =→u' =3 v ' =cosx →v =sen x

(5)

m)

sen

2

x dx

=

sen x · senx dx I = u= sen xu ' =cosx v '=sen x →v =−cosx

sen x cos x +

cos x · cos x dx=−sen x cos x +

cos2x dx=

=−sen x cos x +

(1−sen2x)dx=−sen x cosx +

dx−

sen2x dx

I

→ 2I=−senx cos x +x → I=−1

2sen x cos x + x 2+C

n)

arctan x dx

= u =arctanxu '= 1 √1+x2 v '=1v= x x · arctanx −

x

1+ x2dx t =√=1 +x2 dt = x1+x2dx

x · arctan x−

dt=x · arctan x−t =x · arctan x−

1+ x2+C

o)

arccos x dx

= u =arccos x →u '=− 1 √1−x2 v ' =1v =x x · arccos x +

x

1−x2dx t =√=1− x2 dt =− x1 −x2dx

x ·arccos x −

dt =x · arccosx −t =x ·arccos x−

1− x2+C

p)

x

ln x

2

dx

= u=ln2 x →u '=2 lnxx v ' =xv =1 2x 2 1 2x 2ln2x −

x ln x dx = u =lnxu ' =1x v '= xv=1 2x 2 1 2x 2ln2x−1 2x 2ln x +1 2

x d x = =1 2 x 2ln2x∣−1 2 x 2ln∣x∣+1 4x 2 +C

q)

x

1

2

e

x

dx

= u =( x+1)2 u ' =2(x +1) v '=exv=ex (x +1)2ex

2(x +1)exdx = u =2(x +1) →u ' =2 v '=exv =ex (x +1)2ex−2(x +1)ex+

2exdx=

=(x +1)2ex−2( x+1)ex+2ex=(x2+1)ex+C

r)

x

2

– 3 x

2

ln x dx

= u=ln xu ' =1 x v ' =x2−3 x+ 2 →v =1 3x 3 −3 2x 2+2 x

(

1 3x 3− 3 2x 2+2 x

)

ln x−

(

1 3x 2− 3 2x+2

)

dx =

(

1 3x 3− 3 2x 2+2 x

)

ln∣x∣− 1 9x 3+ 3 4x 22 x +C

s)

ln

1

x

1

x

3

dx

=

(1+ x)− 3 2ln(1+ x)dx = u=ln(1+x )u '= 11+x v ' =(1+ x)− 3 2 v =−2 (1+x )− 1 2 −2(1+x )− 1 2ln(1+x )+2

(1+ x)− 3 2dx= =−2(1+x )− 1 2ln(1+x )−4(1+x )− 1 2=−ln(1+ x )2+4

1+ x +C

4. Integrales racionales:

a)

1

x

2

x –6

dx

=

1 (x +3)(x−2)dx=

(

A x+3+ B x−2

)

dx=

A(x−2)+B(x +3) (x +3)(x −2) dx x=2 → 5 B=1= x=−3 → −5 A=1 =

−1 5· 1 x+3+ 1 5· 1 x−2dx=− 1 5ln( x+3)+ 1 5ln(x −2)= 1 5·ln

(

x −2 x +3

)

+C

b)

3 x

3

x

2

4

dx

3 x3:( x2−4)=3x (Resto : 12 x) →

(

3 x+ 12x (x +2)(x−2)

)

dx=

(

3 x + Ax +2+ Bx −2

)

dx = =

(

3 x +A( x+2)+B(x −2) (x +2)(x−2)

)

dx x= 2 → 4A= 24= x=−2 → −4 B=−24

(

3x + 6 x+2+ 6 x −2

)

dx= 3 2x 2+6ln(x +2)+6ln (x−2)=3 x2 2 +ln

(

x 24

)

6+C

c)

1

x

3

– 4x

2

– 25 x

100

dx

=

1 (x− 4)(x +5 )(x−5)dx =

(

A x−4+ B x+5+ C x−5

)

dx =

A(x +5)(x −5 )+B (x−4)(x−5)+C (x−4 )(x +5 ) (x −4)(x +5)(x −5 ) dx x=4 → −9 A=1= x =−5 → 90B =1 x=5 → 10 C=1 =

(

−1 9· 1x−4+ 190· 1x +5+ 110· 1x−5

)

dx =− 19ln∣x −4∣+ 190ln∣x +5∣+ 110ln∣x − 5∣+ C

(6)

d)

x

2

x

2

x

2

dx

=

x 2+x−2−x +2 x2+x −2 dx=

(

1+ −x +2 (x +2)(x−1)

)

dx=

(

1+ Ax +2+ Bx−1

)

dx= =

(

1+A(x −1)+B( x+2) (x+2)(x−1)

)

dx x =−2 → −3 A=4= x=1 → 3 B=1

(

1− 4 3· 1x+2+ 13· 1x−1

)

dx= =x −4 3ln(x +2)+ 1 3ln(x−1)= x+ 1 3ln

x−1 (x +2)4

+C

e)

x

3

– 2x

2

x – 1

x

2

– 3 x

2

dx

(x3−2 x2+x−1): (x2−3x +2)=x +1 (Resto : 2 x−3) →

(

x +1+ 2 x−3 (x−2)(x−1)

)

dx= =

(

x+1+ A x −2+ B x −1

)

dx=

(

x+1+ A( x−1)+B (x−2) (x−2)(x−1)

)

dx x=1 → −B =−1= x=2 → A=1 =

(

x+1+ 1 x −2+ 1x −1

)

dx= 12x 2 +x +ln(x −2)+ln(x−1)= x2 2+x +ln

x 23 x+ 2

+C

f)

1

x

3

+

x

2

– x−1

dx

=

1 (x +1)2(x−1)dx=

(

A x +1+ B (x+1)2+ C x−1

)

dx =

A(x +1)(x−1)+B(x−1)+C(x +1)2 (x+1)2(x−1) dx x =1 → 4 C=1= x =−1 → −2 B=1 x=0 → − A−B +C=1 =

(

−1 4· 1 x +1− 1 2· 1 (x +1)2+ 1 4· 1 x −1

)

dx=− 1 4ln(x+1)+ 1 2· 1 x+1+ 1 4ln( x−1)= 1 2( x +1)+ 1 4ln

x−1 x+1

+C

g)

2x – 4

x

1

2

x

3

dx

=

(

A x−1+ B (x −1)2+ C x +3

)

dx=

A(x−1)(x+3)+B(x +3)+C (x−1)2 (x−1)2(x+3) dx x=1 → 4 B=−2= x=−3 → 16 C=−10 x =0 → −3 A+ 3 B+ C=−4 =

(

5 8· 1 x −1− 1 2· 1 (x−1)2− 5 8· 1 x+3

)

dx= 5 8ln( x−1)+ 1 2· 1 x−1− 5 8ln(x +3)= 1 2( x− 1)+ 5 8ln

x− 1 x +3

+C

h)

1

x

1



x

3

2

dx

=

(

A x−1+ B x +3+ C (x +3)2

)

dx=

A(x +3)2+B(x−1)(x+3)+C(x−1) (x −1)(x+3)2 dx x=1 → 16 A=1= x=−3 → −4 C=1 x=0 → 9 A−3 B−C=1 =

(

1 16· 1 x−1− 1 16· 1 x+3− 1 4· 1 (x +3)2

)

dx = 1 16ln( x−1)− 1 16· ln(x+3)+ 1 4· 1 x +3= 1 4( x +3)+ 1 16ln

x− 1 x +3

+C

i)

3 x−2

x

2

−4

dx

=

3x−2 (x +2)(x−2)dx=

(

A x+2+ B x−2

)

dx=

A(x−2)+B (x+2) (x +2)(x −2) dx x =2 → 4 B=4= x =−2 → − 4 A=−8 =

(

2 x +2+ 1 x −2

)

dx=2ln(x+2)+ln(x −2 )=ln

(x+ 2) 2 (x− 2)

+C

j)

2x

2

5 x –1

x

3

x

2

– 2 x

dx

=

2 x 2+5 x−1 x (x−1)( x+2)dx =

(

A x+ B x −1+ C x +2

)

dx=

A(x−1)(x +2)+B x (x+2)+C x (x−1) x (x−1)( x+2) dx x =0 → −2 A=−1= x=1 → 3 B=6 x =−2 → 6C=−3 =

(

1 2· 1 x+ 2 x −1− 1 2· 1 x+2

)

dx= 1 2ln x+2ln(x−1)− 1 2ln(x +2)=2ln∣x −1∣+ 1 2ln

x x+ 2

+C

k)

x

2

3

– 3 x

dx

=

−3 x (x−3)dx=

(

A x+ B x−3

)

dx=

A(x−3)+B x x (x−3) dx x =0 → −3 A=−3= x =3 → 3 B=−3 =

(

1 x− 1 x−3

)

dx =ln x −ln( x−3)=ln

x x− 3

+C

(7)

l)

3 x

2

x

3

x

dx

=

−3 x+2 x (x +1)(x−1)dx =

(

A x+ B x +1+ C x−1

)

dx =

A(x +1)(x−1)+B x (x−1)+C x (x +1) x (x +1)(x−1) dx x=0 → −A=2= x=1 → 2 C=−1 x=−1 → 2 B=5 =

(

− 2 x+ 52· 1x +1− 12· 1x−1

)

dx=−2 ln∣x∣+ 52ln∣x +1∣− 12ln∣x −1∣+C

m)

4 x

2

16 x

16

x

2

– 4

2

dx

=

−4x 2+16 x +16 (x+2)2(x−2)2 dx=

(

A x+2+ B (x +2)2+ C x −2+ D (x −2)2

)

dx= =

A(x +2)(x−2) 2+B (x−2)2+C(x +2)2(x−2)+D(x +2)2 (x+2)2 (x−2)2 dx x= 2 → 16 D=32= x=−2 → 16B =−32 x=0 → 8 A+4 B−8 C+4D=16 x =1 → 3 A+ B−9C+ 9 D=28 =

(

1 x+2− 2 (x+2)2− 1 x−2+ 2 (x−2)2

)

dx=ln( x+2)+ 2 x+2−ln(x−2)− 2 x−2=− 8 x24+ln

x+ 2 x− 2

+C

n)

5 x

2

x

3

– 3 x

2

3 x – 1

dx

=

5x 2 (x−1)3dx=

(

A x−1+ B (x−1)2+ C (x−1)3

)

dx= =

A(x−1) 2+B(x −1)+C (x−1)3 dx x=1 → C=5= x =0 → A−B+C =0 x=−1 → 4 A−2B +C=5 =

(

5 x−1+ 10 (x −1)2+ 5 (x −1)3

)

dx=5ln(x−1)− 10 x −1− 5 2· 1 (x−1)2=− 20 x−15 2( x−1)2+5ln∣x−1∣+C

o)

x

4

2 x – 6

x

3

x

2

– 2 x

dx

(x4+2x−6):(x3+x2−2 x)= x−1 (Resto : 3 x26) →

(

x−1+ 3 x 2−6 x (x +2)(x −1)

)

dx= =

(

x−1+A x+ B x +2+ C x−1

)

dx=

(

x−1+ A( x+2)(x−1)+Bx(x −1)+Cx (x+2) x (x +2)(x −1)

)

dx x=0 → −2 A=−6= x =−2 → 6 B=6 x =1 → 3 C=−3 =

(

x−1+3 x+ 1 x +2− 1 x −1

)

dx= 1 2x 2x+3ln x +ln(x +2)−ln(x−1)=x2 2x +ln

x4+2 x3 x −1

+C

p)

2 x – 3

x

3

– 2 x

2

– 9 x

18

dx

=

2 x−3 (x−2)(x +3)(x−3)dx =

(

A x −2+ B x +3+ C x −3

)

dx =

A(x+3)(x−3)+B(x−2)(x−3)+C (x−2)(x +3) (x−2)(x +3)(x −3 ) dx x =2 → −5A=1= x=−3 → 30 B=−9 x =3 → 6 C=3 =

(

−1 5· 1x−2− 310· 1x +3+ 12· 1x−3

)

dx =− 15ln∣x −2∣− 310ln∣x +3∣+ 12ln∣x −3∣+C

q)

1

x

3

+

x

dx

=

1 x (x2+1)dx =

(

A x+ B x+C x2+1

)

dx =

A(x2+1)+Bx2+C x x (x2+1) dx =

(A+B)x2+C x+ A x (x2+1) dx =A+ B=0 C =0 A=1 =

(

1 xx x2+1

)

dx =ln∣x∣− 1 2ln( x 2+1)+C

r)

x

2

x

+

3

+

x+1

x

dx

=

x 2+x+1 x (x2+1)dx =

(

A x+ B x+C x2+1

)

dx =

A(x2+1)+Bx2+C x x (x2+1) dx =

(A+B)x2+C x+ A x (x2+1) dx =A+ B=1 C =1 A=1 =

(

1 x+ 1 x2+1

)

dx =ln∣x∣+arctan x +C

(8)

s)

x+3

x

2

+6 x+10

dx

=

x +3 x2+6 x +9+1dx=

x+3 (x +3)2+1dx =t = x +3 dt=dx

t t2+1dt =12ln(t 2+1)=1 2ln

(

(x +3) 2+1

)

+C

t)

x

1− x

3

+

x

dx

=

1−x x (x2+1)dx =

(

A x+ B x+C x2+1

)

dx =

A(x2+1)+Bx2+C x x (x2+1) dx =

(A+B)x2+C x+ A x (x2+1) dx =A+ B=0 C=−1 A=1 =

(

1 xx +1 x2+1

)

dx =

(

1 xx x2+1− 1 x2+1

)

dx =ln∣x∣− 1 2ln( x 2+1)− arctan x + C

u)

8 x

2

+

x

(

x−2)( 4 x

2

+1)

dx

=

(

A x −2+ B x +C 4 x2+1

)

dx=

A(4 x2+1)+Bx (x−2)+C (x−2) (x−2)(4 x2+1) dx=

(4 A+B)x2+(C−2B)x +A−2C x (x2+1) dx 4 A+ B=8= C−2B =1 A−2 C=0 =

(

2 x −2+ 1 4 x2+1

)

dx=

2 x −2dx+

1 (2 x)2+1dx

t =2x → dt=2 dx =

2 x−2dx + 1 2

1 t2+1dt=ln( x− 2) 2+1 2arctan(2 x )+C

v)

2

x

4

−1

dx

=

2 (x +1)(x−1)( x2+1)dx=

(

A x +1+ B x−1+ C x +D x2+1

)

dx= =

A(x−1)(x 2+1)+B(x +1)(x2+1)+C x (x +1)(x−1)+D(x+1)(x−1) (x+1)(x −1)(x2+1) dx x=1 → 4 B=2= x =−1 → −4 A=2 x =0 → − A+B−D=2 x =2 → 5 A+ 15B + 6C+3D=2 =

(

−1 2· 1 x +1+ 1 2· 1 x−1− 1 x2+1

)

dx= 1 2ln

x−1 x+ 1

arctan x+ C

w)

3 x+ 4

x

3

– 3 x

2

+4 x−12

dx

=

3x +4 (x−3)(x2+4)dx

(

A x−3+ B x+C x2+4

)

dx =

A(x2+4)+Bx( x−3)+C (x−3) (x −2 )(4x2+1) dx =

(A+B) x2+(C −3B) x+4A −3C (x−2)( x2+4) dx = = A+B =0 C −3 B=3 4 A−3C =4

(

x−31 − x x2+4

)

dx=

1 x −3dx −

x x2+4dx

t =x2+4 → dt =2 x dx =

1 x −3dx − 1 2

1 tdt=ln∣x− 3∣− 1 2ln( x 2+4 )+C

x)

3 x

2

+

x+10

x

3

– 2 x

2

+4 x−8

dx

=

3 x 2+x +10 (x−2)(x2+4)dx

(

A x −2+ B x +C x2+4

)

dx =

A(x2+4)+Bx (x−2)+C (x−2) (x −2)(x2+4) dx =

(A+B) x2+(C −2B)x +4A−2C (x−2)(x2+4) dx= = A+B=3 C −2 B=1 4 A−2 C =10

(

x−23 + 1 x2+4

)

dx =

3 x−2dx +

1 4

(

(

x 2

)

2 +1

)

dx

t= x 2 → d t = 12dx =

3 x−2dx + 1 2

1 t2+1dt=3 ln∣x− 2∣+ 1 2arctan

(

x 2

)

+C

y)

x

x

23

+6 x+3

+

x

2

– 2

dx

=

x 2+6 x +3 (x−1)(x2+2 x+2)dx

(

A x−1+ B x +C x2+2x +2

)

dx=

A(x2+2x +2)+Bx (x −1)+C(x −1) (x −1)(x2+2 x+2) dx= =

(A+B) x 2+(2 A−B+C) x+2A−C (x −2)(x2+4) dx A+B =1= 2 A−B+ C=6 2 A−C=3

(

x−12 + −x+1 x2+2x +2

)

dx =

2 x−1dx+

x+1 x2+2x +1+1dx= =

2 x−1dx+

−(x +1)+2 (x+1)2+1dx

t =x +1 → dt =dx =

2 x−1dx−

t t2+1dt +

2 t2+1dt=2ln∣x−1∣− 1 2ln

(

(x+1) 2+1

)

+2arctan( x+1 )+ C

(9)

z)

3 x – 3

x

2

– x +2

dx

3 x –3 x2– 2 ·1 2· x + 1 4− 1 4+2 dx =

3 x –3 2− 3 2

(

x –1 2

)

2 +7 4 dx=

3

(

x –1 2

)

− 3 2 7 4

[

(

2

7

(

x – 1 2

)

)

2 +1

]

dx = t = 2 √7

(

x− 12

)

dt= 2 √7dx =

3 t − 3

7 7 t2+1 dt =

3 t t2+1dt −

3

7 7 t2+1dt = 3 2ln

(

4 7

(

x− 1 2

)

2 +1

)

3

7 7 arctan

(

2

7

(

x − 1 2

)

)

+C

5. Integrales trigonométricas:

a)

cos

7

x dx

=

(

cos2x

)

3cos x dx =

(

1−sen2x

)

3cos x dx

t =sen x → dt = cosx dx =

(1−t2)3dt =

(1−3 t2+3t4t6)dt = =t −t3+3 5t 51 7t 7+C=sen x −sen3x + 3 5sen 5x − 1 7sen 7x +C

b)

cotg x dx

=

cos x sen xdx

t = sen x → dt =cos x dx =

1 tdt =lnt +C=ln( sen x)+ C

c)

3 sen x cos x dx

= t =sen x dt=cos x dx

3t dt=3 2t 2+C=3 2sen 2x + C

d)

4 sen

3

x cos

2

x

=

4sen2cos2x sen x dx =

4

(

1−cos2x

)

cos2x sen x dx

t =cosx → dt =−sen x dx =−

4(1−t2)t2dt = =

(4t4−4t2)dt=4 5t 54 3t 3+C=4 5cos 5x−4 3cos 3x + C

e)

sen x

cos

2

x

dx

= t =cos x dt=−sen x dx

−1 t2dt =

t −2dt=t−1 +C= 1 cos x+C

f)

sen x – tg x

cos x

dx

=

sen x−sen x cos x cos x dx =

1− 1 cos x cos x sen x dx

t =cos x → dt =−sen xdx =

1 t−1 t dt =

(

1 t2− 1 t

)

dt= =−1 t−ln t +C=− 1 cos xln∣cos x∣+C

g)

sen

5

xcos

3

x dx

=

sen5x cos2x cos x dx=

sen5x (1−sen2x) cos x dx

t =sen x → dt =cosx dx =

t5(1−t2)dt =

(t5−t7)dt = =1 6t 61 8t 8+C=1 6sen 6x − 1 8sen 8x +C

h)

cos

3

x

1– sen x

dx

=

cos 2x 1−sen xcos x dx=

1−sen2x 1−sen x cos x dx

t =sen x dt= cosx dx =

1−t 2 1−t dt =

(1+t )(1−t ) 1−t dt = =

(1+t )dt =t +1 2t 2+C=sen x +1 2sen 2x +C

i)

1

dx

sen x

=

1−sen x (1+sen x)(1−sen x )dx =

1−sen x 1−sen2xdx=

1−sen x cos2x dx=

1 cos2xd x +

sen x cos2x dx

t =cos x dt =−sen x dx = =tan x +

1 t2dt =tan x− 1 t+C=tan x− 1

cos x+C=tan x−sec x +C =

sen x−1

(10)

j)

dx

sen

2

xcos

2

x

=

(

1 sen2x+ 1 cos2x

)

dx=

1 sen2xdx

t= π 2−x → dt =−dx +

1 cos2xdx=

−1 sen2

(

π2−t

)

dt

sen(π2t)=cost +

1 cos2xdx=

−1 cos2tdt +

1 cos2xdx=

=−tant +tan x+C=−tan

(

π

2−x

)

+tan x +C=tan x −

1 tan x+C=

sen x−cos x

sen x cos x +C

También con el cambio t = tan x → d t= dx

cos2x sen 2x =1−cos2x =1− 1 1+tan2x= tan2x 1+ tan2x= t2 1+t2

sen2d xx cos2xt =tan x=

1+t2 t2 dt=

(

1 t2+1

)

dt= −1 t +t +C=tan x− 1 tan x+C

6. Determina las siguientes integrales por el método que consideres más conveniente:

a)

x

1

3

x

dx

= x=t6 dx =6t5d t

∫ √

t6 1+3

t66t 5dt =

6 t8 1+t2dt

t8: (t2+1)=t6 −t4 +t2−1 (Resto: 1) =6

(

t6t4+t2−1+ 1 1+t2

)

dt = =6 7t 76 5t 5+6 3t 3−6t +6 arctan(t )+C=6 7 6

x76 5 6

x5+2

x−66

x +6 arctan

6x +C

b)

ln

1

x

dx

=

ln

(

x−1

)

dx=

−ln x dx = u =ln(x) → u '= 1x v '=−1 → v =−xx ln x +

1dx=x−ln x +C=x (1−ln∣x∣)+C

c)

x

1 ln

x

1

dx

= u =ln(x+ 1) → u ' = 1x +1 v '=x +1 → v =2 3(x +1) 3 2 2 3(x+1) 3 2ln(x +1)−2 3

x +1dx= 2 3(x+1) 3 2ln∣x +1∣−4 9(x + 1) 3 2+C

d)

dx

x

1

ln

2

x

= t =ln x dt =dx x

1+tdt2=arctan(t )+C=arctan(ln∣x∣)+ C

e)

cos(ln x)dx

cos (ln x )dx I = u=cos(lnx ) → u '=− 1xsen(ln x) v '=1 → v= x x cos(ln x)+

sen(ln x)dx = u =sen(lnx ) → u '= 1xcos(ln x) v '=1 → v=x

=x cos(ln x)+ x sen(ln x)−

cos(ln x)

I

→ 2I=x cos(ln x)+ x sen(ln x)+C → I =x

2

(

cos(ln∣x∣)+ sen (ln∣x∣)

)

f)

1 – x

2

dx

=

x=sent → dx =cost dt

1− x2

=√1−sen2t =cos2t =cost

cos2t dt

I = u =cost → u '=−sent v ' =cost → v =sen t

sent cost +

sen2t dt =

=sen t cost +

(

1−cos2t

)

dt =sent cost +

dt −I → 2I =sent cost +t +C I=1 2sen t cost + t 2+C= 1 2x cos(arcsen x)+ arcsen x 2 +C= x

1− x2 2 + arcsen x 2 +C

g)

x

x

1

dx

= t =x +1 → dt= 12 dxx +1 x =t2−1

2(t2−1)dt =2 3t 3−2t +C=2 3(x +1) 3 22

x +1+C

h)

dx

x

x

1

t=x +1 → dt =1= 2 dxx +1 x=t2−1

t2−12 dt=

(

A t +1+ B t−1

)

dt =

A(t −1)+B(t +1) (t +1)(t −1) dt x=1 → 2 B=2= x =−1 → −2 A=2 =

(

1 t−1− 1t +1

)

dt=ln(t−1)−ln(t +1)+C=ln

x +1−1

x+ 1+1

+C

(11)

i)

x

5

e

x3

dx

=

x3ex3 x2d x = t =x3 → dt =3 x2dx

1 3tetdt = u= 1 3t → u ' =13 v ' =et → v=−et −1 3t et +1 3

etdt= =−1 3t et1 3et+C=−1 3

(

1+ x 3

)

ex3 +C

j)

e

x

e

2x

3 e

x

2

dx

= t=ex → dt =ex dx

1 t2+3t +2dt=

(

A t +1+ B t +2

)

dt =

A(t +2)+B(t +1) (t +1)(t +2) dt x =−1 → A=1= x=−2 → −B=1 =

(

1 t +1− 1 t +2

)

dt =ln(t +1)−ln(t +2)+C=ln

(

ex+1 ex+2

)

+C

7. Calcula la función f(x) que cumple que f''(x) = 6x+1, f(0)=1 y f(1)=0.

f ' (x)=

f ' ' (x )dx =3 x2 +x +C f (x )=

f ' ( x)dx= x3 +1 2x 2 +C x +D

{

x =0 → f (0)=D=1 x =1 → f (1)=1+1 2+C+1=0 → f (x )= x3+1 2x 25 2x+ 1

8. Encuentra la familia de funciones que tienen como segunda derivada f''(x)=4x sabiendo que

tienen un máximo relativo en x=-1.

Determina cuál de las funciones de la familia pasa por el punto

1,

5

6

.

f ' (x )=

f ' ' (x)dx =2x2+C f (x )=

f ' (x )dx=2 3x 3 +C x +D

{

f ' (−1)=0 →Máximo 2+C=0 Familia: f ( x )=2 3 x 32 x +D → Por el punto

(

1,− 5 6

)

f ( x)= 2 3x 32 x +1 2

9. Busca una primitiva F(x) de la función f(x)=2x – 4 que verifique que

F

x

≥

0

x

∈[

0, 4

]

.

F ( x)=

(2 x – 4)dx= x2– 4 x+C Si hacemos C=4 tendremos: F ( x )= x24 x + 4=(x−2)2≥0∀ x ∈ℝ

10.Halla f(x) sabiendo que:

f ' '

x

=

cos

x

2

f '

2 

=

0

f(0)=1

f ' (x)=

cos

(

x 2

)

dx=2sen

(

x 2

)

+C f (x )=

[

2sen

(

x 2

)

+C

]

dx=−4cos

(

x 2

)

+Cx+D

{

f ' (2 π)=0 → C=0 f (0)=1 → −4+D=1 → f ( x)=−4cos

(

x 2

)

+5

11.Calcula el área encerrada entre la gráfica de las funciones

f x=x , g x=2 x , hx=x

2

.

A=

0 1

(

g(x )−f ( x)

)

dx +

1 2

(

g(x )−h(x)

)

dx =

0 1 (2 x−x )dx +

1 2

(

2x −x2

)

dx= =

[

1 2x 2

]

0 1 +

[

x2−1 3x 3

]

1 2 =1 2+4− 8 3−1+ 1 3= 7 6u.d.s.

12.Integrales definidas:

a)

0 

sen2x dx

=

[

−1 2cos 2x

]

0 π =−1 2cos2 π+ 1 2cos0=0

b)

1 2

x

2

ln xdx

u =lnx → u '= 1= x v '=x2 → v =1 3x 3

[

1 3x 3 ln x

]

1 2 −

1 2 x2 3dx= 8 3ln 2−0−

[

x3 9

]

1 2 =8 3ln 2− 8 9+ 1 9= 24 ln(2)− 7 9

c)

−1 3

∣2 x

4∣dx

= ∣2 x− 4∣=

{

4−2 x si x ≤2 2 x− 4 si x >2

−1 2 (4−2 x)dx +

2 3 (2 x−4)dx =

[

4x −x2

]

−1 2 +

[

x2−4 x

]

2 3 =4+5−3+4=10

(12)

d)

−1 0

x

1

x dx

u =x → u ' =1= v ' =1+ x → v= 2 3( 1+x ) 3 2

[

2 3x (1+x ) 3 2

]

−1 0 −

1 2 2 3(1+x ) 3 2dx=0−

[

4 15(1+x ) 5 2

]

−1 0 =− 4 15

e)

−3 3

x

1∣dx

= ∣x −1∣=

{

1− x si x≤1x−1 si x>1

−3 1 (1−x)dx +

1 3 (x−1)dx=

[

x −1 2x 2

]

−3 1 +

[

1 2x 2x

]

1 3 =1 2+ 15 2+ 3 2+ 1 2=10

f)

0 1

x dx

=

[

2 3x 3 2

]

0 1 =2 3−0= 2 3

g)

1 e

ln x

x

dx

= t =lnx → d t = 1 xdx x =1→t =0 x =e→t =1

0 1 t dt =

[

1 2t 2

]

0 1 =1 2−0= 1 2

h)

0 2

x

x

2

1

dx

= t =x2+1 → d t =2 x dx x=0→t =1 x =2→t =5

1 5 1 2

tdt=

[√

t

]

1 5 =

5−1

i)

1 e e

2 ln x dx

= u =lnx → u '= 1 x v '=2 → v=2 x

[

2x ln x

]

1 e e

1 e e 2dx=2e+2 e

[

2x

]

1 e e =2e+2 e−2e+ 2 e= 4 e

j)

−3 3

x

2

x

2

1

dx

función= par 2

0 √3 x2 x2+1dx=2

0 √3 x2+1−1 x2+1 dx=2

0 √3

(

1− 1 1+x2

)

dx=2

[

x −arctan x

]

0√ 3 =2

(

3− π 3

)

13.Dadas la parábola

y

=

x

2

1

y la recta y=5 – x, representa y calcula el área de los

siguientes recintos:

a) Recinto acotado limitado por ambas curvas.

Intersección de funciones: x2−1=5−x → x2+x−6=0 → x 1=−3 x2=2 A=

−3 2

(

(5−x )−( x2−1)

)

dx=

−3 2 (−x2x +6)dx=

[

−1 3x 31 2x 2+6 x

]

−3 2 =22 3 −

(

− 27 2

)

= 125 6 u.d.s.

b) Recinto acotado limitado por la parábola a la izquierda, la recta a la

derecha y el eje OX por debajo.

A=

1 2 (x2−1)dx +

2 5 (5− x)dx=

[

1 3x 3x

]

1 2 +

[

5x −1 2x 2

]

2 5 =2 3+ 2 3+ 25 2−8= 35 6 u.d.s.

c) Recinto acotado limitado por el eje OY a la izquierda, la parábola a la

derecha, la recta por encima y el eje OX por debajo.

A=

0 1 (5−x )dx +

1 2

[

(5− x)−(x2−1)

]

dx=

[

5 x−1 2x 2

]

0 1 +

[

−1 3x 31 2x 2+6x

]

1 2 =9 2+ 22 3− 31 6= 20 3 u.d.s.

14. Calcula el área encerrada entre las gráficas de las siguientes funciones en cada caso:

a)

f

x

=

x

2

– 5 g

x

=−

x

2

5

Intersección de funciones: x2−5=−x2+5 → 2x2−10=0 → x=±

5 A=

−√5 √5

(

(5− x2)−(x2−5)

)

dx = Función par 2

0 √5 (10− 2x2)dx=2

[

10 x−2 3x 3

]

0 √5 =20

5−4

125 3 u.d.s.

b)

f

x

=

4 – x

2

g

x

=

8 – 2 x

2 Intersección de funciones: 4−x2=8−2x2 → x2=4 → x =±2 A=

−2 2

(

(8−2 x2)−(4−x2)

)

dx = Función par 2

0 2 (4−x2)dx=2

[

4 x−1 3x 3

]

0 2 =32 3 u.d.s.

(13)

c)

f ( x)= x( x−3) g( x)=2 x−4

Intersección de funciones: x (x−3)=2 x− 4 → x2−5 x +4=0 → x 1=1 x2=4 A=

1 4

(

(2x−4)−x (x−3)

)

dx=

1 4

(

x2+5 x−4

)

dx=

[

−1 3x 3+5 2x 2−4 x

]

1 4 =9 2 u.d.s.

d)

f

x

=

6

x

g

x

=

7

x

Intersección de funciones:6 x=7−x → x 2−7 x+6=0 → x 1=1 x2=6 A=

1 6

(

7−x−6 x

)

dx=

[

7x − 1 2x 2−6ln x

]

1 6 =35 26ln6 u.d.s.

e)

f

x

=

sen x g

x

=

cos x

con 0

x

2 

Intersección de funciones:sen x=cos x → tanx=1 → x1= π 4

45º x2=5 π 4

235º A=

π 4 5 π 4

(sen x −cosx ) dx=

[

−cos x−sen x

]

π 4 5 π 4=2

2 u.d.s.

f)

f

x

=

x

2

g

x

=

∣1

x∣

Intersección de funciones:

x 2=∣1−x∣ → x 2=(1− x) 2 → 2 x2−5 x +2=0 → x 1= 1 2 x2=2 A=

1 2 2

(

x 2−∣(1− x)∣

)

dx ∣1−x∣=

{

1− x si x ≤1x −1 si x>1=

1 2 1

(

x 2−(1− x)

)

dx+

1 2

(

x 2−(x−1)

)

dx= =

[

2x

x 3

2 −x + 1 2x 2

]

1 2 1 +

[

2 x

x 3

2 − 1 2x 2+x

]

1 2 =13 24 u.d.s.

g)

f

x

=

x

3

2x g

x

=

x

2 Intersección de funciones: x3−2 x= x2 → x3x2−2 x=0 → x 1=−1 x2=0 x3=2 A=

−1 0

(

(x3−2x )−x2

)

dx+

0 2

(

x2−(x3−2 x)

)

dx=

[

1 4x 41 3x 3x2

]

−1 0 +

[

−1 4x 4+1 3x 3+x2

]

0 2 =37 12 u.d.s.

15.Halla el área del recinto acotado limitado por las gráficas de las funciones

f

x

=

e

x2

,

g

x

=

e

x

y el eje OY

Intersección:ex+2=ex → x +2=−x → x =−1 A=

−1 0

(

ex +2ex

)

dx=

[

ex +2+ex

]

−1 0 =e2+1− 2e u.d.s.

16.Considera el área de la región encerrada entre la parábola

y

=

x

2

y la recta y=1.

Esta región se divide mediante una recta horizontal y=a.

Determina a que hace que dicha región quede dividida en mitades de igual superficie.

Intersección con y=1: x2=1 → x=±1 Intersección con y=a : x2=a → x =±a ATOTAL=

−1 1

(

1−x2

)

dx=

[

x −1 3x 3

]

−1 1 =4

3u.d.s. Área bajo y=a: A(a)=

a

a (a −x2)dx=

[

ax−1 3x 3

]

−√aa =2a

a−2 3(

a) 3 =4 3a 3 2 4 3a 3 2 = 1 2ATOTAL 2 3 → a 3 2=1 2 → a= 1 3

4

Figure

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