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Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 25 maths textbook solution

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Answer: \frac{-1}{2(x+\sin x)^{2}}+c

Hint:Use substitution method to solve this integral.

Given:   \int \frac{1+\cos x}{(x+\sin x)^{3}} d x


        \begin{aligned} &\text { Let } I=\int \frac{1+\cos x}{(x+\sin x)^{2}} d x \\ &\text { Put } x+\sin x=t \Rightarrow(1+\cos x) d x=d t \text { then } \end{aligned}

        I=\int \frac{1}{t^{3}} d t=\int t^{-3} d t=\frac{t^{-3+1}}{-3+1}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]

        =\frac{t^{-2}}{-2}+c=\frac{-1}{2} \frac{1}{t^{2}}+c

        =\frac{-1}{2(x+\sin x)^{2}}+c \quad[\because t=x+\sin x]


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