Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.26 question 16 maths

explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.26 question 16 maths

Answers (1)

Answer:
The correct answer is e^{x}(\log x)+e^{x}\left(\frac{-1}{x}\right)+c
Hint:

\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+c

Given:

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

Solution:

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

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

\int e^{x}\left(\log x+\frac{1}{x}\right) d x+\int e^{x}\left(-\frac{1}{x}+\frac{1}{x^{2}}\right) d x

We have,

\begin{aligned} &\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+c \\ &\int e^{x}\left(\log x+\frac{1}{x}\right) d x+\int e^{x}\left(-\frac{1}{x}+\frac{1}{x^{2}}\right) d x \end{aligned}

            =e^{x}(\log x)+e^{x}\left(\frac{-1}{x}\right)+c

So, the correct answer is   e^{x}(\log x)+e^{x}\left(\frac{-1}{x}\right)+c

Posted by

infoexpert26

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

CBSE Date Sheet 2027 LIVE: Class 10, 12 exam dates announcement soon at cbse.gov.in, sample papers out
anam-khan

CBSE Board Exams 2027: LOC submission for Classes 10, 12 begins tomorrow; check dates and fees
anam-khan

CBSE Date Sheet 2027 LIVE: Class 10, 12 exam dates soon; LOC submission begins, guidelines

Latest Question