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

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

Answers (1)

Answer:

\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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