Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 10 maths textbook solution.

Please solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 10 maths textbook solution.

Answers (1)

Answer:- \frac{\pi}{4}

Hints:-  You must know about the integration rules of functions .

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

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

We can write integrand using partial fractions

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

Cancelling denominators

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

Putting x=1, \text { in (1) }

            \begin{aligned} &x=(A x+B)(x+1)+C\left(x^{2}+1\right) \\ &1=(A \times 1+B)(1+1)+C\left(1^{2}+1\right) \\ &1=2(A+B)+C(2) \\ &1=2 A+2 B+2 C \end{aligned}                  .....(2)

Putting x=0 \text { in (1) }

            \begin{aligned} &x=(A x+B)(x+1)+C\left(x^{2}+1\right) \\ &0=(A \times 0+B)(0+1)+C(0+1) \end{aligned}

           \begin{aligned} &0=B+C \\ &B=-C \end{aligned}

Put in (2)

           \begin{aligned} &1=2 A-2 C+2 C \\ &A=\frac{1}{2} \end{aligned}

Putting x = -1

           \begin{aligned} &-1=\left(A(-1)+B(-1+1)+C\left((-1)^{2}+1\right)\right. \\ &-1=(-A+B)(0)+C(2) \\ &-1=2 C \\ &C=\frac{-1}{2} \end{aligned}

Put value of A and C in (2)

           \begin{aligned} &2 A+2 B+2 C=1 \\ &2 \times \frac{1}{2}+2 B+2 \times \frac{-1}{2}=1 \\ &1+2 B-1=1 \\ &2 B=1 \\ &B=\frac{1}{2} \end{aligned}

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

Hence we can write

      \begin{gathered} \frac{x}{(x+1)\left(x^{2}+1\right)}=\frac{\left(\frac{1}{2} x+\frac{1}{2}\right)}{\left(x^{2}+1\right)}+\frac{\frac{-1}{2}}{(x+1)} \\ =\frac{x}{2\left(x^{2}+1\right)}+\frac{1}{2\left(x^{2}+1\right)}-\frac{1}{2(x+1)} \end{gathered}

Solving

  \begin{aligned} &I_{3}=\frac{-1}{2} \int_{0}^{\infty} \frac{1}{x+1} d x \\ &=\frac{-1}{2}\left[\log |x+1|_{0}^{\infty}\right. \\ &=\frac{-1}{2}[\log (1)-\log (1) \mid] \\ &=0 \end{aligned}

Hence,

           \begin{aligned} &I=I_{1}+I_{2}+I_{3} \\ &=0+\frac{\pi}{4}+0 \\ &=\frac{\pi}{4} \end{aligned}

Posted by

infoexpert23

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 Class 12 result verification 2024 application out at cbse.gov.in; last date, fees
anam-khan

CBSE results 2024 later for 102 students over fake certificates; board issues show cause notice
anam-khan

What after CBSE Class 12 results 2024? From CUET UG to JEE Advanced, all you need to know

Latest Question