integration help!!!!!

Question

∫xdx/(x+1)(x^2+1)

Answer 1

Use  partial  fractions

Let  x  =  -1

Let  x = 0

0  =  A + B

Substitute  in A  value

Integrate  the  sum  term  by  term

Expanding  the  Integrate

Where s = 1+ x

 ds  =  dx

Substitute  s  and  u   values

The  solution  is

 

 

 

Answer 2

 ∫ xdx/(x+1)(x^2+1)

1/[(x + 1)(x^2 + 1)] = A/(x + 1) + (Bx + C)/(x^2 + 1)

=> 1 = A(x^2 + 1) + (x + 1)(Bx + C).

i) Let x = -1 => 2A = 1 => A = 1/2

ii) Let x = i => (C + Bi)(1 + i) = 1 => (C - B) + (C + B)i = 1.

For part ii), comparing real and imaginary parts gives the equation:

C - B = 1 and C + B = 0.

This implies that C = 1/2 and B = -1/2.

1/[(x + 1)(x^2 + 1)] = (1/2)/(x + 1) + [(-1/2)x + 1/2]/(x^2 + 1)

=> 1/[(x + 1)(x^2 + 1)] = 1/[2(x + 1)] - (x - 1)/[2(x^2 + 1)].

Therefore:

=∫ 1/[(x + 1)(x^2 + 1)] dx

= ∫ {1/[2(x + 1)] - (x - 1)/[2(x^2 + 1)]} dx

= 1/2 ∫ 1/(x + 1) dx - 1/2 ∫ (x - 1)/(x^2 + 1) dx

= 1/2 ∫ 1/(x + 1) dx - 1/2 ∫ [x/(x^2 + 1) - 1/(x^2 + 1)] dx

= 1/2 ∫ 1/(x + 1) dx - 1/2 ∫ x/(x^2 + 1) dx + 1/2 ∫ 1/(x^2 + 1) dx

= 1/2 ∫ 1/(x + 1) dx - 1/4 ∫ 2x/(x^2 + 1) dx + 1/2 ∫ 1/(x^2 + 1) dx

= (1/2)log|x + 1| - (1/4)log(x^2 + 1) + (1/2)arctan(x)