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Explain solution RD Sharma class 12 Chapter 19 Definite Integrals Exercise Revision Exercise question 15

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Answer:  \frac{\pi}{4}

Given:  \int_{0}^{\infty} \frac{x}{(1+x)\left(1+x^{2}\right)} d x

Hint: Use Partial fraction method


\int_{0}^{\infty} \frac{x}{(1+x)\left(1+x^{2}\right)} d x

Using Partial fraction

\begin{aligned} &\frac{x}{(1+x)\left(1+x^{2}\right)}=\frac{A}{1+x}+\frac{B x+c}{1+x^{2}} \\ & \end{aligned}

\Rightarrow x=A\left(1+x^{2}\right)+(B x+C)(1+x) \\

\Rightarrow x=A+A x^{2}+B x+B x^{2}+C+C x \\

\Rightarrow x=(A+C)+(A+B) x^{2}+(B+C) x

On comparing

\begin{aligned} &A+C=0, B+C=1, A+B=0 \\ & \end{aligned}

\therefore A=\frac{-1}{2}, B=\frac{1}{2}, C=\frac{1}{2}


\begin{aligned} &\int_{0}^{\infty} \frac{x}{(1+x)\left(1+x^{2}\right)} d x=\int_{0}^{\infty}\left(\frac{-1}{2(1+x)}+\frac{1}{2}\left(\frac{x+1}{1+x^{2}}\right)\right) d x \\ & \end{aligned}

=-\frac{1}{2}[\log |1+x|]_{0}^{\infty}+\frac{1}{2} \int_{0}^{\infty} \frac{x}{1+x^{2}}+\frac{1}{1+x^{2}} d x

\begin{aligned} &=-\frac{1}{2}\left[\log \frac{\sqrt{1+x^{2}}}{x+1}\right]_{0}^{\infty}+\frac{1}{2}\left(\tan ^{-1} x\right)_{0}^{\infty} \\ & \end{aligned}

=\frac{1}{2} \times 0+\frac{1}{2}\left(\frac{\pi}{2}-0\right)=\frac{\pi}{4}

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