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Question Number 110549 by shahria14 last updated on 29/Aug/20

Commented by peter frank last updated on 29/Aug/20

Answered by peter frank last updated on 29/Aug/20

$$\mathrm{an}\stackrel{}{\mathrm{s}}=\frac{{\mathrm{e}}^{\mathrm{x}}}{1+\mathrm{x}}$$

Answered by Dwaipayan Shikari last updated on 29/Aug/20

$$\int \frac{{xe}^x+e^x}{{(1+x)}^2}-\frac{e^x}{{(1+x)}^2}dx$$$$\int e^x(\frac{1}{1+x}-\frac{1}{{(1+x)}^2})dx=\frac{e^x}{1+x}+C$$

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