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

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

Answers (1)

Answer:  \frac{\pi}{4}

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

Hint: Use Partial fraction method

Solution:

\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}

Hence,

\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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