The value of int_{0}^{infty} frac{x}{(1+x)(x^ | vedclass

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

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$\frac{\pi}{4}$

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(B) Let $I = \int_{0}^{\infty} \frac{x}{(1+x)(x^{2}+1)} dx$.Using partial fractions,we write $\frac{x}{(1+x)(x^{2}+1)} = \frac{A}{1+x} + \frac{Bx+C}{x^{2}+1}$.Equating numerators: $x = A(x^{2}+1) + (Bx+C)(1+x) = (A+B)x^{2} + (B+C)x + (A+C)$.Comparing coefficients: $A+B=0$,$B+C=1$,$A+C=0$.Solving these,we get $A = -\frac{1}{2}$,$B = \frac{1}{2}$,and $C = \frac{1}{2}$.Thus,$I = \int_{0}^{\infty} \left( -\frac{1}{2(1+x)} + \frac{x+1}{2(x^{2}+1)} ight) dx = -\frac{1}{2} \int_{0}^{\infty} \frac{dx}{1+x} + \frac{1}{4} \int_{0}^{\infty} \frac{2x}{x^{2}+1} dx + \frac{1}{2} \int_{0}^{\infty} \frac{dx}{x^{2}+1}$.Evaluating the integrals: $I = \left[ -\frac{1}{2} \ln(1+x) + \frac{1}{4} \ln(x^{2}+1) + \frac{1}{2} 	an^{-1}(x) ight]_{0}^{\infty}$.Combining the logarithmic terms: $I = \left[ \frac{1}{4} \ln \left( \frac{(x^{2}+1)}{(1+x)^{2}} ight) + \frac{1}{2} 	an^{-1}(x) ight]_{0}^{\infty}$.As $x 	o \infty$,$\frac{x^{2}+1}{(1+x)^{2}} 	o 1$,so $\ln(1) = 0$.At $x=0$,$\frac{1}{4} \ln(1) + \frac{1}{2} 	an^{-1}(0) = 0$.Therefore,$I = 0 + \frac{1}{2} \left( \frac{\pi}{2} - 0 ight) = \frac{\pi}{4}$.

The value of $\int_{0}^{\infty} \frac{x}{(1+x)(x^{2}+1)} dx$ is

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