Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions Question 6 Maths Textbook Solution.

Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions Question 6 Maths Textbook Solution.

Answers (1)

Answer:

                                                                                \tan ^{-1}\left(\log _{e} x\right)^{2}+C

Given:

\int \frac{1}{1+\left(\log _{e} x\right)^{2}} d x \quad \text { w.r.t. }\left(\log _{e} x\right)

Hint:

Using \int \frac{1}{1+t^{2}} d x

Explanation:

Let I=\int \frac{1}{1+\left(\log _{e} x\right)^{2}} d x(\log x)                                            

           =\int \frac{1}{1+\left(\log _{e} x\right)^{2}} \frac{d x}{x}                                                        \left[\frac{d}{d x}(\log x)=\frac{1}{x} \Rightarrow d(\log x)=\frac{d x}{x}\right]

            =\int \frac{1}{1+t^{2}} \cdot d t                                                                   \text { [Put } \left.\log _{e} x=t \Rightarrow \frac{1}{x} d x=d t\right]

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

             =\tan ^{-1}\left(\log _{e} x\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

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