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 18 Indefinite Integrals Excercise 18.14 Question 4

Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.14 Question 4

Answers (1)

Answer: x-\frac{5}{2} \tan ^{-1}\left|\frac{x}{2}\right|+c

Hint: To solve this integral, use special integral formula.

Given:  \int \frac{x^{2}-1}{x^{2}+4} d x

Solution:

Let,

\begin{aligned} &I=\int \frac{x^{2}-1}{x^{2}+4} d x=\int \frac{x^{2}-1+4-4}{x^{2}+4} d x\\ \\ &=\int \frac{\left(x^{2}+4\right)-1-4}{x^{2}+4} d x=\int\left(\frac{x^{2}+4}{x^{2}+4}-\frac{5}{x^{2}+4}\right) d x \\\\ &=\int\left(1-\frac{5}{x^{2}+4}\right) d x=\int 1 d x-5 \int \frac{1}{x^{2}+2^{2}} \end{aligned}

 

=x-5 \cdot \frac{1}{2} \tan ^{-1}\left(\frac{x}{2}\right)+c \quad\quad\quad\quad\quad\quad\left[\because \int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left|\frac{x}{a}\right|+c\right]

 

=x-\frac{5}{2} \tan ^{-1}\left|\frac{x}{2}\right|+c

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

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