Let us assume
I = \(\int\limits_{0}^{\infty}\cfrac{\text x}{(1+\text x)(1+\text x^2)}d\text x \)
Adding – 1 and + 1

Let I1 = \(\int\limits_{0}^{\infty}\cfrac{1}{(1+\text x^2)}d\text x \)
I2 = \(\int\limits_{0}^{\infty}\cfrac{\text x}{(1+\text x)(1+\text x^2)}d\text x \)
Thus I = I1 – I2 …….equation 1
Solving for I1

Let


a + b = 0; a + c = 1; b + c = 0
solving we get
a = c = 1/2
b = – 1/2
substituting the values in equation 3

Solving:

Thus

Substituting values equation 2 and equation 4 in equation 1
Thus
