Question

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

Answer 1

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

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

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

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

Let $f\left(x\right)=\frac{1}{x+1}$

$\Rightarrow f\text{'}\left(x\right)=\frac{-1}{{\left(x+1\right)}^2}$

$\therefore I=e^x\left(\frac{1}{x+1}\right)+C\left[\int e^x\left[f\left(x\right)+f\text{'}\left(x\right)\right]dx=e^{x}f\left(x\right)+C\right]$


$\int e^x\frac{x}{{(x+1)}^2}dx=$ $\underline{e^x\left(\frac{1}{x+1}\right)+C}$.