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Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.26 question 13 maths textbook solution

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Answer:
The correct answer is e^{x} \cdot \frac{1}{x+2}+c
Hint:

\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+c

Given:

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

Solution:

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

        \begin{aligned} &\mathrm{I}=\int e^{x}\left[\frac{x+1+1-1}{(x+2)^{2}}\right] d x \\ &\mathrm{I}=\int e^{x}\left[\frac{x+2}{(x+2)^{2}}-\frac{1}{(x+2)^{2}}\right] d x \\ &\mathrm{I}=\int e^{x}\left[\frac{1}{(x+2)}-\frac{1}{(x+2)^{2}}\right] d x \end{aligned}

\text { let } f(x)=\frac{1}{(x+2)} \text { and } f^{\prime}(x)=-\frac{1}{(x+2)^{2}}

        \begin{aligned} &\mathrm{I}=\int e^{x}\left[f(x)+f^{\prime}(x)\right] d x=e^{x} f(x)+c \\ &\mathrm{I}=e^{x} \frac{1}{x+2}+c \end{aligned}

 

 

 

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