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 Indefinite Integrals exercise 18.9 question 67

need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.9 question 67

Answers (1)

best_answer

Answer: \log \left|\frac{x+1}{\sqrt{x^{2}+2 x+2}}\right|+c

Hint: Use substitution method to solve this integral

Given: \int \frac{1}{(x+1)\left(x^{2}+2 x+2\right)} d x

Solution:

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

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

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

        \text { Putting } x+1=\tan u        ......(i)

        \Rightarrow d x=\sec ^{2} u \; d u \text { then }

        I=\int \frac{1}{\tan u\left\{\tan ^{2} u+1\right\}} \sec ^{2} u\; d u

           =\int \frac{1}{\tan u\left\{\sec ^{2} u\right\}} \sec ^{2} u \; d u\left[\because 1+\tan ^{2} u=\sec ^{2} u\right]

           =\int \frac{1}{\tan u} d u=\int \cot u\; d u

           =\log |\sin u|+c            ........(ii)     \left[\because \int \cot x \; d x=\log |\sin x|+c\right]

        \text { Also, from (i) } x+1=\tan u=\frac{\sin u}{\cos u} \quad\left[\because \tan x=\frac{\sin x}{\cos x}\right]

        \begin{aligned} &\Rightarrow(x+1) \cos u=\sin u \\ &\Rightarrow(x+1)^{2} \cos ^{2} u=\sin ^{2} u \quad[\text { Squaring on both sides }] \\ &\Rightarrow(x+1)^{2}\left(1-\sin ^{2} u\right)=\sin ^{2} u \quad\left[\because \cos ^{2} x=1-\sin ^{2} x\right] \\ &\Rightarrow(x+1)^{2}-(x+1)^{2} \sin ^{2} u=\sin ^{2} u \\ &\Rightarrow(x+1)^{2}=\sin ^{2} u+(x+1)^{2} \sin ^{2} u \\ &\Rightarrow(x+1)^{2}=\left[1+(x+1)^{2}\right] \sin ^{2} u \end{aligned}

        \Rightarrow \sin ^{2} u=\frac{(x+1)^{2}}{1+(x+1)^{2}}=\frac{(x+1)^{2}}{x^{2}+2 x+1+1}=\frac{(x+1)^{2}}{x^{2}+2 x+2} \quad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right]

        \Rightarrow \sin u=\sqrt{\frac{(x+1)^{2}}{x^{2}+2 x+2}}=\frac{(x+1)}{\sqrt{x^{2}+2 x+2}}    ........(iii)

        \begin{aligned} &\text { From (ii) and (iii) we get }\\ &I=\log \left|\frac{x+1}{\sqrt{x^{2}+2 x+2}}\right|+c \end{aligned}

 

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 Board exams twice a year from 2025, no plan for semesters: MoE
anam-khan

CBSE Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th
anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years

Latest Question