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 18 Indefinite Integrals exercise 18.30 question 21

explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 21

Answers (1)

best_answer

Answer:

           \frac{1}{6} \log \left|\frac{(1+x)^{6}(-1+x)^{2}}{(2 x+1)^{5}}\right|+C

Hint:

            To solve this integration, we use partial fraction method   

Given:

            \int \frac{x^{2}+1}{(2 x+1)\left(x^{2}-1\right)} d x \\

Explanation:

Let

\begin{aligned} &I=\int \frac{x^{2}+1}{(2 x+1)\left(x^{2}-1\right)} d x \\ &I=\int \frac{x^{2}+1}{(2 x+1)(x-1)(x+1)} d x \ldots\left[x^{2}-y^{2}=(x+y)(x-y)\right] \\ &\frac{x^{2}+1}{(2 x+1)(x-1)(x+1)}=\frac{A}{2 x+1}+\frac{B}{x-1}+\frac{C}{x+1} \\ &x^{2}+1=A(x-1)(x+1)+B(2 x+1)(x+1)+C(x-1)(2 x+1) \end{aligned}

\begin{aligned} &\text { At } x=1 \\ &1+1=A(0)+B(3)(2)+C(0) \\ &2=6 B \\ &B=\frac{1}{3} \end{aligned}
\begin{aligned} &B=\frac{1}{3} \\ &\text { At } x=-1 \\ &1+1=A(0)+B(0)+C(-1)(-2) \\ &2=2 C \\ &C=1 \end{aligned}
\begin{aligned} &\text { At } x=\frac{-1}{2} \\ &\frac{1}{4}+1=A\left(-\frac{1}{2}+1\right)\left(-\frac{1}{2}-1\right)+B(0)+C(0) \\ &\frac{5}{4}=A\left(\frac{-3}{4}\right) \\ &A=\frac{-5}{3} \end{aligned}
\begin{aligned} &\frac{x^{2}+1}{(2 x+1)(x-1)(x+1)}=\frac{-5}{3(2 x+1)}+\frac{1}{3(x-1)}+\frac{1}{(x+1)} \\ &I=\frac{-5}{3} \int \frac{1}{2 x-1} d x+\frac{1}{3} \int \frac{1}{x-1} d x+\int \frac{1}{x+1} d x \end{aligned}
\begin{aligned} &I=\frac{1}{3}\left[\frac{-5}{2} \log |2 x+1|+\log |x-1|+3 \log |x+1|\right]+C \\ &I=\frac{1}{6}[-5 \log |2 x+1|+2 \log |x-1|+6 \log |x+1|]+C \\ &I=\frac{1}{6} \log \left|\frac{(x-1)^{2}(x+1)^{6}}{(2 x+1)^{5}}\right|+C \end{aligned}

 

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 admit card 2025 out for Class 10, 12 supplementary exams
anam-khan

CBSE Class 10, 12 practical exams 2025 for compartment students from July 10; detailed guidelines issued
anam-khan

CBSE Central Sector Scheme of Scholarship 2025 applications open on scholarships.gov.in

Latest Question