Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.16 question 11

Need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.16 question 11

Answers (1)

Answer:

        \frac{1}{6} \log \left|\frac{x^{6}}{x^{6}+1}\right|+C

Hint:

Use substitution method as well as special integration formula to solve this type of problem

Given:

        \int \frac{1}{x\left(x^{6}+1\right)} d x

Solution:

Let\: \: I=\int \frac{1}{x\left(x^{6}+1\right)} d x

                \begin{aligned} &=\int \frac{x^{5}}{x^{5}\left\{x\left(x^{6}+1\right)\right\}} d x \\ &=\int \frac{x^{5}}{x^{6}\left(x^{6}+1\right)} d x \end{aligned}

Put\: \: x^{6}=t\Rightarrow 6x^{5}dx=dt\Rightarrow x^{5}dx=\frac{dt}{6}

Then\: \: I=\int \frac{1}{t\left(t+1\right)} \frac{dt}{6}

                \begin{aligned} &=\frac{1}{6} \int \frac{1}{t^{2}+t} d t \\ &=\frac{1}{6} \int \frac{1}{t^{2}+2 \cdot t \cdot \frac{1}{2}+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}} d t \\ &=\frac{1}{6} \int \frac{1}{\left(t+\frac{1}{2}\right)^{2}-\frac{1}{4}} d t \end{aligned}

Put\: \: t+\frac{1}{2}=u\Rightarrow dt=du

                \begin{aligned} =\frac{1}{6} \int \frac{1}{u^{2}-(\frac{1}{2})^{2}} d t \end{aligned}\begin{aligned} =\frac{1}{6} \int \frac{1}{u^{2}-(\frac{1}{2})^{2}} d u \end{aligned}

                \begin{aligned} &=\frac{1}{6} \cdot \frac{1}{2 \times\left(\frac{1}{2}\right)} \log \left|\frac{u-\frac{1}{2}}{u+\frac{1}{2}}\right|+C \quad\left[\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\right] \\ &=\frac{1}{6} \log \left|\frac{\frac{2u-1}{2}}{\frac{2u+1}{2}}\right|+C \\ &=\frac{1}{6} \log \left|\frac{2 u-1}{2 u+1}\right|+C \end{aligned}

                \begin{aligned} &=\frac{1}{6} \log \left|\frac{2\left(t+\frac{1}{2}\right)-1}{2\left(t+\frac{1}{2}\right)+1}\right|+C \quad\left[\because t+\frac{1}{2}=u\right] \\ &=\frac{1}{6} \log \left|\frac{2 t+1-1}{2 t+1+1}\right|+C \\ &=\frac{1}{6} \log \left|\frac{2 t}{2 t+2}\right|+C \\ &=\frac{1}{6} \log \left|\frac{t}{t+1}\right|+C \\ &=\frac{1}{6} \log \left|\frac{x^{6}}{x^{6}+1}\right|+C \quad\left[\because t=x^{6}\right] \end{aligned}

Posted by

Gurleen Kaur

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 English Paper Analysis: Questions easy, ‘student-friendly’; can 'score above 90%'
anam-khan

CBSE Class 12 Board Exam 2026 LIVE: 12th English paper easy, 'competency-based questions' for literature
anam-khan

QR code in CBSE Class 12 maths paper redirects to 'rickroll' prank video; board says exam integrity intact

Latest Question