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 18.15 Question 5

Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.15 Question 5

Answers (1)

Answer:

            \frac{1}{2} \tan ^{-1}\left(\frac{x+3}{2}\right)+C

Hint:

            To solve this problem use special integration formula

Given:

            \int \frac{1}{x^{2}+6 x+13} d x

Solution:

Let   I=\int \frac{1}{x^{2}+6 x+13} d x

            \begin{aligned} &=\int \frac{1}{x^{2}+2 \cdot x \cdot 3+3^{2}-3^{2}+13} d x \\ & \end{aligned}

           =\int \frac{1}{(x+3)^{2}-9+13} d x \\

           =\int \frac{1}{(x+3)^{2}+4} d x

Put    x+3=t \Rightarrow d x=d t

Then   I=\int \frac{1}{t^{2}+2^{2}} d t

            =\frac{1}{2} \tan ^{-1}\left(\frac{t}{2}\right)+C                               \left[\because \int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\right]

            =\frac{1}{2} \tan ^{-1}\left(\frac{x+3}{2}\right)+C                          \quad[\because t=x+3]

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

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14
anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens
anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question