$\int_{0}^{\infty} \frac{x}{(1 + x)(1 + x^2)} dx$ equals to :

Question

Professor PC Thomas Classes · Answer

Let $I = \int_{0}^{\infty} \frac{x}{(1+x)(1+x^2)} dx$. Substitute $x = 1/t$, then $dx = -1/t^2 dt$. The integral becomes $I = \int_{\infty}^{0} \frac{1/t}{(1+1/t)(1+1/t^2)} \left(-\frac{1}{t^2}ight) dt = \int_{0}^{\infty} \frac{t^3}{t(t+1)(t^2+1)} \frac{1}{t^2} dt = \int_{0}^{\infty} \frac{1}{(1+t)(1+t^2)} dt$ (Option C is correct). Now add both expressions for $I$: $2I = \int_{0}^{\infty} \frac{x+1}{(x+1)(x^2+1)} dx = \int_{0}^{\infty} \frac{1}{1+x^2} dx = \left[ 	an^{-1} x ight]_0^\infty = \frac{\pi}{2}$. Therefore, $I = \frac{\pi}{4}$ (Option A is correct).