Ejercicios resueltos Integrales.pdf

 e ln(x) 1 x

´  sin(x) cos(x) dx

´ 

u  = sin(x)

dx

→ du = cos(x) dx

ˆ 

  sin(x) cos(x) dx = =

ˆ 

udu

u2

+ cte 2 sin2 (x) = + cte 2

u  = sin(x) dv  = cos(x)

ˆ 

→ →

du  = cos(x)dx v  = sin(x)

  sin(x) cos(x) dx   = sin2(x)

−

ˆ 

  sin(x) cos(x) dx

ˆ 

2   sin(x) cos(x) dx   = sin2 (x)

sin2 (x)   sin(x) cos(x) dx = + cte 2

ˆ 

u  = ln(x)

ˆ  ln(x) x

→

dx =

=

1

du  =

ˆ 

x

dx

u du

u2

+ cte 2 ln2 (x) = + cte 2

e

ˆ  ln(x) x

ln2 (x) dx = 2

e 1

|

1

ln2 (e) = 2 1 = 2

u  = ln(x) dv  =

1 x

→ →

−

du  =

 ln 2 (1) 2

1 x

 dx

v  = ln(x)

e

ˆ  ln(x) x

e

dx = ln2 (x)

1 e

2

ˆ  ln(x) x

dx

e 1

| −

ˆ  ln(x) x

1

= ln2 (x)

|

= ln2 (e)

− ln (1)

1

dx

e 1 2

= 1

e

ˆ  ln(x) x

1 2

dx =

1

2

´ [(x − 1) (x + 1)]− x−1 u  = x+1

u  =

u

x 1 x + 1

− → du =

ˆ 

[(x2

1.(x+1)−(x−1),1 (x+1)2

− 1) (x + 1)]−

2 3

dx =

= =

dx  =

2 3

dx

  −  = 3

x 1 x+1

→ 21 du  = (x +dx1)

2  dx (x+1)2

ˆ  [(x − 1) (x + 1) (x + 1)]− ˆ   − dx (x − 1)(x + 1) ˆ  − (x − 1) (x + 1) − dx ˆ  (x − 1)− dx ˆ  (xx +− 1)1  − dx ˆ  −x +1 1 (x + 1) 2

2 3

2

4

3

3

2 3

=

4 3

= =

2 3

2

u

2 3

2

 du

1 3

1u = + cte 2 13 3 = 2



x 1 x + 1

−



1 3

+ cte

2 3

dx

2

u =

1 3

  − 1  − 1  =  + 1  + 1   1 − 1 −  1 (  + 1) − ( − 1) 1  3  + 1 (  + 1)  − 1 − 2 1 3 − +1 1− (  + 1) 3

x x

x x

2

du = du =

x x

. x

3

x

,

2

x

dx

2 3

x x

2

x

 dx

2

3 du = 2

ˆ 

[(x2

x x + 1

3

dx

(x + 1) 2

− 1) (x + 1)]−

2 3

dx =

=

ˆ   x − 1  − ˆ  3 x + 1 2

dx

2 3

(x + 1) 2

 du

3  u  +  cte 2 3 x 1 = 2 x + 1 =



−



1 3

+ cte

x  = 0 x

f (x) =

ˆ 

(1 +  t)3 ln(1 + t) dt

0

f   (0) 2 f   (0) 3   P 3 (x) =  f   (0) + f  (0) x + x + x 2! 3!

0

f   (0) =

ˆ 

(1 +  t)3 ln(1 + t) dt  = 0

0

f 

(1 +  x)3 ln(1 + x)

x>

f   (x) = (1 + x)3 ln(1 + x)

−1 3

→ f   (0) = (1 + 0)

ln(1 + 0) = 1,0 = 0

1  = 3(1 + x)2 ln(1 + x) + (1 + x)2 1 +  x 2 2   f  (0) = 3(1 + 0) ln(1 + 0) + (1 + 0) = 3,1,0 + 1 = 1 1  + 2(1 + x) = 6(1 + x) ln(1 + x) + 5(1 + x) f   (x) = 6(1 + x) ln(1 + x) + 3(1 + x)2 1 +  x f   (0) = 6(1 + 0) ln(1 + 0) + 5(1 + 0) = 6,1,0 + 5 = 5 f   (x) = 3(1 + x)2 ln(1 + x) + (1 + x)3

P 3 (x) = 0 + 0x +

1 2 5 3 x + x 2! 3!

1 2 5 3 x + x 2 6 1 5 = x2 +  x 2 6 =





2x

ˆ 

F (x) = √ 

arctan(t) dt

x

0

F (x) =

2x

ˆ 

√ x arctan(t) dt +

0

ˆ 

ˆ √ 

x

ˆ 

√ x arctan(t) dt

ˆ √ 

=

−

−

arctan(t) dt

0

x

F (x) =

arctan(t) dt

0

2x

arctan(t) dt +

0

f (x) = arctan(x)

ˆ 

arctan(t) dt

0

R

F  (x) =

=

− arctan(√ x) √ x + arctan(2x) (2x) 1  + arctan(2x) 2 − arctan(√ x) 2√  x √  1  arctan( x) + 2 arctan(2x) − 2√  x

F  (x) =

y  =

√  y  = x √ x = →

1 4

x2

− x = 0

√ x; y  =

A  = y  = 21 x 1 x 2

 9 1 a 2

´  (

x

1 x 2

; x  = 9

− √ x) dx

a

√ 

( x)2 = ( 12 x)2

→ x =

1 4

x2

x  = 0 y x  = 4 a  = 4 9

A

= = = =

A =

− √  − |  − − −

1 x x)dx 4 2 1 2 2 3/2 9 x x 4 4 3 81 16 18 (4 ) 4 3 81 16 + 22 4 3 43 12

ˆ 



(

−