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  • 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)

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}

 

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