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Question Number 217431 by peter frank last updated on 13/Mar/25

Answered by Frix last updated on 13/Mar/25

$$\mathrm{Simply}\mathrm{by}\mathrm{parts}:$$$$u^{\prime }=\frac{1}{{(x+1)}^2}\to u=-\frac{1}{x+1}$$$$v=x{\mathrm{e}}^x\to v^{\prime }=(x+1){\mathrm{e}}^x$$$$\int u^{\prime }v=uv-\int {uv}^{\prime }$$$$\int \frac{x{\mathrm{e}}^x}{{(x+1)}^2}=-\frac{x{\mathrm{e}}^x}{x+1}+\int {\mathrm{e}}^{x}dx=...=\frac{{\mathrm{e}}^x}{x+1}+C$$

Answered by Spillover last updated on 13/Mar/25

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