Question

Simplify: $\int \frac{e^x\left(x-1\right)}{{\left(x+1\right)}^3}dx$

Answer 1

$I=\int e^x\frac{(x-1)}{(x+1{)}^3}dx$

$I=\int e^x\frac{(x+1-2)}{(x+1{)}^3}dx$

$I=\int e^x\frac{1+x}{(x+1{)}^3}dx$ $-$ $\int e^x\frac{2}{(1+x{)}^3}dx$

$I=\int e^x\frac{1}{(1+x{)}^2}dx$ $-$ $\int e^x\frac{2}{(x+1{)}^3}dx$

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

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

$I=e^x\frac{1}{(1+x{)}^2}$ $+$ $\int e^x\frac{2}{(1+x{)}^3}dx$ $-$ $\int e^x\frac{2}{(1+x{)}^3}dx$

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