Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Provide Solution For R D Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18 point 19 Question 2 Maths Textbook Solution

Provide Solution For R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.18 Question 2 Maths Textbook Solution.

Answers (1)

Answer: \frac{1}{2} \log \left|x^{2}+x+3\right|+\frac{1}{\sqrt{11}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{11}}\right)+c

Hint: Integrate by ILATE

Given: \int \frac{x+1}{x^{2}+x+3} d x

Solution:

\begin{aligned} &I_{1}=\frac{1}{2} \int \frac{2 x+2}{x^{2}+x+3} d x \\ &I_{1}=\frac{1}{2} \int \frac{(2 x+1)+1}{x^{2}+x+3} d x \end{aligned}

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

\begin{aligned} I_{1} &=\frac{1}{2} \int \frac{2 x+2}{x^{2}+x+3} d x \\ I_{1} &=\frac{1}{2} \int \frac{(2 x+1)+1}{x^{2}+x+3} d x \\ &=\frac{1}{2} \int \frac{2 x+1}{x^{2}+x+3} d x+\frac{1}{2} \int \frac{1}{x^{2}+x+3} d x \end{aligned}                                                                                            \left[\int \frac{d t}{t}=\ln t+c,\right]

                          Let,x^{2}+x+3=t

               (2 x+1) d x=d t

                            I_{1}=\log \left|x^{2}+x+3\right|+\mathrm{C}

                I_{2}=\int \frac{1}{x^{2}+x+3} d x

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

                    =\int \frac{1}{\left(x+\frac{1}{2}\right)^{2}+\frac{11}{4}} d x                                                        \left[\begin{array}{c} x+\frac{1}{2}=t \Rightarrow d x=d t \\ \sqrt{\frac{11}{4}}=a=\frac{\sqrt{11}}{2} \end{array}\right]

                    =\frac{1}{\frac{\sqrt{11}}{2}} \tan ^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{11}}{2}}\right)+C_{2}                                    \left[\int \frac{1}{t^{2}+a^{2}} d t=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)+C_{2}\right]

            I_{2}=\frac{2}{\sqrt{11}} \tan ^{-1} \frac{(2 x+1)}{\sqrt{11}}+C_{2}

          I=\frac{1}{2}\left(I_{1}+I_{2}\right)

                         =\frac{1}{2}\left[\log \left|x^{2}+x+3\right|+\frac{2}{\sqrt{11}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{11}}\right)\right]+c

                         =\frac{1}{2} \log \left|x^{2}+x+3\right|+\frac{1}{\sqrt{11}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{11}}\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

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