Evaluate the definite integral: int_{0}^{inft | vedclass

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Question

(A) Let $I = \int_{0}^{\infty} \frac{x}{(1 + x)(1 + x^2)} \, dx$. Using the property $\int_{0}^{\infty} f(x) \, dx = \int_{0}^{\infty} f(1/x) \cdot \f

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(A) Let $I = \int_{0}^{\infty} \frac{x}{(1 + x)(1 + x^2)} \, dx$. Using the property $\int_{0}^{\infty} f(x) \, dx = \int_{0}^{\infty} f(1/x) \cdot \frac{1}{x^2} \, dx$,we substitute $x = 1/t$,so $dx = -1/t^2 \, dt$. As $x 	o 0, t 	o \infty$ and as $x 	o \infty, t 	o 0$. $I = \int_{\infty}^{0} \frac{1/t}{(1 + 1/t)(1 + 1/t^2)} \cdot (-1/t^2) \, dt = \int_{0}^{\infty} \frac{1/t}{(\frac{t+1}{t})(\frac{t^2+1}{t^2})} \cdot \frac{1}{t^2} \, dt = \int_{0}^{\infty} \frac{1}{t(t+1)(t^2+1)/t^3} \cdot \frac{1}{t^2} \, dt = \int_{0}^{\infty} \frac{1}{(t+1)(t^2+1)} \, dt$. Now,$2I = \int_{0}^{\infty} \frac{x}{(1+x)(1+x^2)} \, dx + \int_{0}^{\infty} \frac{1}{(1+x)(1+x^2)} \, dx = \int_{0}^{\infty} \frac{x+1}{(1+x)(1+x^2)} \, dx = \int_{0}^{\infty} \frac{1}{1+x^2} \, dx$. $2I = [	an^{-1}(x)]_{0}^{\infty} = \frac{\pi}{2} - 0 = \frac{\pi}{2}$. Therefore,$I = \frac{\pi}{4}$.

Evaluate the definite integral: $\int_{0}^{\infty} \frac{x}{(1 + x)(1 + x^2)} \, dx$

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