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Explain Solution R.D. Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions  Question 43 Maths Textbook Solution.

Answers (1)

Answer:

e^{x} \log (x+1)+C

Given:

\int \frac{e^{x}}{x+1}[1+(x+1) \log (x+1)] d x

Hint:

Using integration by parts and \int e^{x} d x

Explanation:

Let \mathrm{I}=\int \frac{e^{x}}{x+1}[1+(x+1) \log (x+1)] d x

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

=\log (x+1) e^{x}-\int \frac{1}{x+1} e^{x} d x+\int \frac{1}{x+1} e^{x} d x \ldots\left\{\int u \cdot v d x=u \int v d x-\int \frac{d u}{d x} \int v d x d x\right\}

=\log (x+1) e^{x}+C

Hence,\int \frac{e^{x}}{x+1}[1+(x+1) \log (x+1)] d x=\log (x+1) e^{x}+C

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