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 R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 5 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 5 Maths Textbook Solution.

Answers (1)

Answer:

-\frac{1}{x}+\log |x+1|+c

Given:

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

Hint:

Use partial function method.

Solution:  

I=\int \frac{1+x+x^{2}}{x^{2}(1+x)} d x

using partial function

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

 now,

\frac{1+x+x^{2}}{x^{2}(1+x)}=\int \frac{1}{x^{2}}+\frac{1}{1+x} d x

                           =-\frac{1}{x}+\log |x+1|+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 Time Table 2025 Live: UP Board exams from February 24; CBSE, BSEB date sheet soon
anam-khan

CBSE Date Sheet 2025 Live: Class 10, 12 board exam dates soon; syllabus, sample papers
anam-khan

Extra time in CBSE exams for students with type 1 diabetes: Kerala SHRC seeks report on plea

Latest Question