#### Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions  Question 18 Maths Textbook Solution.

$\frac{e^{x}}{x+4}+C$

Given:

$\int \frac{x+3}{(x+4)^{2}} e^{x} d x$

Hint:

Using $\left[\int u . v d x=u \int v d x-\int \frac{d u}{d x} \int v d x d x\right.$

Explanation:

Let $\mathrm{I}=\int \frac{x+3}{(x+4)^{2}} e^{x} d x$

$=\int \frac{(x+3+1-1)}{(x+4)^{2}} e^{x} d x$

$=\int e^{x}\left[\frac{x+4}{(x+4)^{2}}-\frac{1}{(x+4)^{2}}\right] d x$

\begin{aligned} &=\int e^{x}\left[\frac{1}{x+4}-\frac{1}{(x+4)^{2}}\right] d x \\ &=\int e^{x} \frac{1}{x+4} d x-\int e^{x} \frac{1}{(x+4)^{2}} d x \\ &=\frac{1}{x+4} \cdot e^{x}-\int \frac{-1}{(x+4)^{2}} \cdot e^{x} d x-\int \frac{1}{(x+4)^{2}} \cdot e^{x} d x \ldots \ldots \ldots . .\left\{\int u . v d x=u \int v d x-\int \frac{d u}{d x} \int v d x d x\right\} \\ &=\frac{e^{x}}{x+4}+\int \frac{1}{(x+4)^{2}} e^{x} d x-\int \frac{1}{(x+4)^{2}} e^{x} d x \\ &=\frac{e^{x}}{x+4}+C \end{aligned}

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