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 63

need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.9 question 63

Answers (1)

Answer: (x+1)+2 \sqrt{x+1}-2 \tan ^{-1}(\sqrt{x+1})-2 \log |x+2|+c

Hint: Use substitution method to solve this integral

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

Solution:

        \text { Let } \mathrm{I}=\int \frac{x+\sqrt{x+1}}{x+2}

        \text { Put } x+1=t^{2} \Rightarrow d x=2 t\; d t \text { then }

        I=\int \frac{\left(t^{2}-1\right)+\sqrt{t^{2}}}{t^{2}+1} 2 t \; d t=2 \int \frac{\left(t^{2}-1\right)+t}{t^{2}+1} t\; d t \quad\left[\begin{array}{l} \because x+1=t^{2} \\ \Rightarrow x=t^{2}-1 \end{array}\right]

        \Rightarrow I=2 \int\left(\frac{t^{2}+t-1}{t^{2}+1}\right) t \; d t=2 \int\left(\frac{t^{2} \cdot t+t . t-t}{t^{2}+1}\right) d t

        \begin{aligned} &\Rightarrow I=2 \int\left(\frac{t^{3}+t^{2}-t}{t^{2}+1}\right) d t=2 \int\left\{\frac{t^{3}}{t^{2}+1}+\frac{t^{2}}{t^{2}+1}-\frac{t}{t^{2}+1}\right\} d t \\ &\Rightarrow I=2\left[\int \frac{t^{3}}{t^{2}+1} d t+\int \frac{t^{2}}{t^{2}+1} d t-\int \frac{t}{t^{2}+1} d t\right] \end{aligned}

        We can write

        I=2\left(I_{1}+I_{2}-I_{3}\right)           .........(i)

        \begin{aligned} \text { where } I_{1}=\int \frac{t^{3}}{t^{2}+1} d t \end{aligned}

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

            \text { and } I_{3}=\int \frac{t}{t^{2}+1} d t

        \text { Now } I_{1}=\int \frac{t^{3}}{t^{2}+1} d t=\int\left(\frac{t^{3}+t-t}{t^{2}+1}\right) d t

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

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

                       =\int t\; d t-\int \frac{t}{t^{2}+1} d t

        \begin{aligned} &\text { put } t^{2}+1=u \Rightarrow 2 t \; d t=d u \Rightarrow t \; d t=\frac{d u}{2} \text { then } \\ &I_{1}=\int t \; d t-\int \frac{1}{u} \frac{d u}{2}=\int t\; d t-\frac{1}{2} \int \frac{d u}{u} \end{aligned}

             =\frac{t^{1+1}}{1+1}-\frac{1}{2} \log |u|+c_{1}\left[\begin{array}{l} \because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \\ \int \frac{1}{x} d x=\log |x|+c \end{array}\right]

            =\frac{t^{2}}{2}-\frac{1}{2} \log \left|1+t^{2}\right|+c_{1}                   .......(ii)              \left[\because u=t^{2}+1\right]

        \text { And } I_{2}=\int \frac{t^{2}}{t^{2}+1} d t=\int\left(\frac{t^{2}+t-t}{t^{2}+1}\right) d t

                       \begin{aligned} &=\int\left(\frac{\left(t^{2}+1\right)-1}{t^{2}+1}\right) d t=\int\left(\frac{t^{2}+1}{t^{2}+1}-\frac{1}{t^{2}+1}\right) d t \\ &=\int\left(1-\frac{1}{t^{2}+1}\right) d t=\int t^{0} d t-\int \frac{1}{t^{2}+1} d t \end{aligned}

                       =\frac{t^{0+1}}{0+1}-\tan ^{-1}(t)+c_{2}\left[\begin{array}{l} \because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \\ \int \frac{1}{x} d x=\log |x|+c \end{array}\right]

                       =t-\tan ^{-1}(t)+c_{2}        .......(iii)    

           \begin{aligned} &\text { Also } I_{3}=\int \frac{t}{t^{2}+1} d t \\ &\text { put } t^{2}+1=p \Rightarrow 2 t d t=d p \Rightarrow t d t=\frac{d p}{2} \text { then } \\ &I_{3}=\int \frac{1}{p} \frac{d p}{2}=\frac{1}{2} \int \frac{1}{p} d p=\frac{1}{2} \log |p|+c_{3} \end{aligned}

                =\frac{1}{2} \log \left|1+t^{2}\right|+c_{3}              ......(iv)

        \begin{aligned} &\text { Substituting the values of } I_{1}, I_{2}, I_{3} \text { from eqn(ii), (iii) and (iv) in }(i) \text { then }\\ &I=2\left[\frac{t^{2}}{2}-\frac{1}{2} \log \left|1+t^{2}\right|+c_{1}+t-\tan ^{-1}(t)+c_{2}-\frac{1}{2} \log \left|1+t^{2}\right|-c_{3}\right] \end{aligned}    

            =2\left[\frac{t^{2}}{2}+t-\tan ^{-1}(t)-\left(\frac{1}{2}+\frac{1}{2}\right) \log \left|1+t^{2}\right|+c_{1}+c_{2}-c_{3}\right]

            =2\left[\frac{t^{2}}{2}+t-\tan ^{-1}(t)-\log \left|1+t^{2}\right|+c_{4}\right] \quad\left[\because c_{4}=c_{1}+c_{2}-c_{3}\right]

            \begin{aligned} &=2 \cdot \frac{t^{2}}{2}+2 t-2 \tan ^{-1}(t)-2 \log \left|1+t^{2}\right|+2 c_{4} \\ &=t^{2}+2 t-2 \tan ^{-1}(t)-2 \log \left|1+t^{2}\right|+c \end{aligned}

            \begin{aligned} &\text { [ since } x+1=t^{2} \text { ] } \\ &=(x+1)+2 \sqrt{x+1}-2 \tan ^{-1} \sqrt{x+1}-2 \log |1+x+1|+c \\ &\mathrm{I}=(x+1)+2 \sqrt{x+1}-2 \tan ^{-1}(\sqrt{x+1})-2 \log |x+2|+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 12th Results LIVE: Class 12 marksheet on cbse.gov.in, DigiLocker soon; OSM system, passing marks
anam-khan

CBSE 12th Result 2026 Date LIVE: Class 12 results at cbse.gov.in soon; how to check on Digilocker, Umang
anam-khan

CBSE 12th Result 2026 Date: Here’s what previous years’ trends indicate

Latest Question