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Question Number 91302 by 174 last updated on 29/Apr/20

Answered by MJS last updated on 29/Apr/20

$$\int {\mathrm{e}}^x\frac{{(x-1)}^2}{{(x^2+1)}^2}dx=$$$$=\int \frac{{\mathrm{e}}^x}{x^2+1}dx-2\int \frac{{\mathrm{e}}^{x}x}{{(x^2+1)}^2}dx$$$$\mathrm{the}{2}^{\mathrm{nd}}\mathrm{one}\mathrm{by}\mathrm{parts}:$$$$-2\int \frac{{\mathrm{e}}^{x}x}{{(x^2+1)}^2}dx=$$$$u^{\prime }=\frac{x}{{(x^2+1)}^2}\to u=-\frac{1}{2(x^2+1)}$$$$v=v^{\prime }={\mathrm{e}}^x$$$$=\frac{{\mathrm{e}}^x}{x^2+1}-\int \frac{{\mathrm{e}}^x}{x^2+1}dx$$$$\mathrm{so}\mathrm{we}\mathrm{have}$$$$\int \frac{{\mathrm{e}}^x}{x^2+1}dx+\frac{{\mathrm{e}}^x}{x^2+1}-\int \frac{{\mathrm{e}}^x}{x^2+1}dx=$$$$=\frac{{\mathrm{e}}^x}{x^2+1}+\mathrm{C}$$

Answered by $@ty@m123 last updated on 29/Apr/20

$$Formula:$$$$\int e^x\{f\left(x\right)+f{}^{\prime }\left(x\right)\}dx=e^{x}f\left(x\right)+\mathrm{C}$$$$Here,$$$$f\left(x\right)=\frac{1}{1+x^2}$$

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