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  • 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}

 

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