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Question Number 116279 by mohammad17 last updated on 02/Oct/20

$$\int \frac{{xe}^x}{{(x+1)}^2}dx$$

Answered by Dwaipayan Shikari last updated on 02/Oct/20

$$\int \frac{(\mathrm{t}-1){\mathrm{e}}^{(\mathrm{t}-1)}}{{\mathrm{t}}^2}\mathrm{dt}=\frac{1}{\mathrm{e}}\int {\mathrm{e}}^{\mathrm{t}}(\frac{1}{\mathrm{t}}-\frac{1}{{\mathrm{t}}^2})\mathrm{dt}\mathrm{x}+1=\mathrm{t}$$$$=\frac{1}{\mathrm{e}}.{\mathrm{e}}^{\mathrm{t}}.\frac{1}{\mathrm{t}}+\mathrm{C}=\frac{1}{\mathrm{t}}{\mathrm{e}}^{\mathrm{t}-1}+\mathrm{C}=\frac{{\mathrm{e}}^{\mathrm{x}}}{\mathrm{x}+1}+\mathrm{C}$$

Commented by mohammad17 last updated on 02/Oct/20

$$thankyousir$$

Answered by TANMAY PANACEA last updated on 02/Oct/20

$$\int \frac{(x+1)\frac{d}{dx}\left(e^x\right)-e^x\frac{d}{dx}(x+1)}{{(x+1)}^2}dx$$$$\int \frac{d}{dx}\left(\frac{e^x}{x+1}\right)dx=\frac{e^x}{x+1}+c$$

Commented by mohammad17 last updated on 02/Oct/20

$$thankyousir$$

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