Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • 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)

best_answer

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}

Posted by

infoexpert27

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Supplementary Exam 2025: Registrations open for Class 10, 12 private students today
anam-khan

CBSE opens application to access 10th answer scripts; verification, revaluation for Class 12 starts on May 31
anam-khan

CBSE supplementary exams 2025 from July 15; form submission starts on May 30

Latest Question