Question

Evaluate: $\int {\left(\frac{1-x}{1+x^2}\right)}^2{e}^{x}dx$

Answer 1

Consider, $I=\int e^x\frac{(1-x{)}^2}{(1+x^2{)}^2}dx$

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

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

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

By integration by parts :

$\int \left[f\right(x\left)g\right(x\left)\right]dx=f\left(x\right)\cdot \int g\left(x\right)dx-\int \left[f^{\prime }\right(x)\int g(x\left)dx\right]dx$

By integration by parts of first integral

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

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

$I=e^x\frac{1}{(1+x^2)}$ $+$ $C$