int_{0}^{infty} frac{x , dx}{(1 + x)(1 + x^2) | vedclass

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(C) Let $I = \int_{0}^{\infty} \frac{x}{(1 + x)(1 + x^2)} \, dx$. Using partial fractions,we write: $\frac{x}{(1 + x)(1 + x^2)} = \frac{A}{1 + x} + \f

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$\pi / 4$

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(C) Let $I = \int_{0}^{\infty} \frac{x}{(1 + x)(1 + x^2)} \, dx$. Using partial fractions,we write: $\frac{x}{(1 + x)(1 + x^2)} = \frac{A}{1 + x} + \frac{Bx + C}{1 + x^2}$. Solving for constants,we get $A = -1/2$,$B = 1/2$,and $C = 1/2$. Thus,$I = \int_{0}^{\infty} \left( \frac{-1/2}{1 + x} + \frac{1/2 x + 1/2}{1 + x^2} ight) \, dx$. $I = -\frac{1}{2} \int_{0}^{\infty} \frac{dx}{1 + x} + \frac{1}{4} \int_{0}^{\infty} \frac{2x}{1 + x^2} \, dx + \frac{1}{2} \int_{0}^{\infty} \frac{dx}{1 + x^2}$. Evaluating the integral: $I = \left[ -\frac{1}{2} \ln(1 + x) + \frac{1}{4} \ln(1 + x^2) + \frac{1}{2} 	an^{-1}(x) ight]_{0}^{\infty}$. $I = \left[ \frac{1}{4} \ln\left( \frac{1 + x^2}{(1 + x)^2} ight) + \frac{1}{2} 	an^{-1}(x) ight]_{0}^{\infty}$. As $x 	o \infty$,$\frac{1 + x^2}{(1 + x)^2} 	o 1$,so $\ln(1) = 0$. $I = (0 + \frac{1}{2} \cdot \frac{\pi}{2}) - (0 + 0) = \frac{\pi}{4}$.

$\int_{0}^{\infty} \frac{x \, dx}{(1 + x)(1 + x^2)} = $

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