Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 106 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 106 Maths Textbook Solution.

Answers (1)

Answer: I=\frac{1}{3} \log \left|\frac{\sqrt{1+x^{8}}-1}{\sqrt{1+x^{8}+1}}\right|+c

Hint: to solve this equation we have to do with differentiate method.

Given: \int \frac{1}{x \sqrt{1+x^{3}}} d x

Solution:

x^{3}=t

3 x^{2} d x=d t

x^{2} d x=\frac{1}{3} d t

I=\int \frac{x^{2}}{x^{3} \sqrt{1+x^{3}}} d x

I=\frac{1}{3} \int \frac{d t}{t \sqrt{1+t}}

\text { Let } 1+\mathrm{t}=\mathrm{p}^{2} \text { , } \mathrm{dt}=2 \mathrm{pdp}

I=\frac{1}{3} \int \frac{2 p d p}{\left.p^{2}-1\right) p}

I=\frac{2}{3} \int \frac{d p}{\left(p^{2}-1\right)}

\frac{2}{3} \times \frac{1}{3} \times \frac{1}{2} \log \left[\frac{p-1}{p+1}\right]+C \quad\left[\frac{d x}{x^{2}-a^{2}}=\frac{1}{2 a} \log \left[\frac{x-a}{x+a}\right]+C\right]

I=\frac{1}{3} \log \left|\frac{\sqrt{1+x^{3}}-1}{\sqrt{1+x^{3}}+1}\right|+c

 

 

Posted by

infoexpert21

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

Board Exam Result 2025 Live: UK Board 10th, 12th results tomorrow; AP SSC, TS Inter, UP, CBSE updates
anam-khan

When will CBSE Class 12 results 2025 be declared? Last two years' trends
anam-khan

CBSE Class 10, 12 board exams 2025 from April 7 to 11 for sports students

Latest Question