Question

Calcule a integral por partes

Answer 1

$I=\displaystyle\int\!\frac{x\cdot e^x}{(x+1)^2}\,dx$


Método de integração por partes:

$\begin{array}{lcl} u=x\,e^x&~\Rightarrow~&du=(e^x+xe^x)\,dx=(x+1)\,e^x\,dx\\\\ dv=\dfrac{1}{(x+1)^2}&~\Leftarrow~&v=-\,\dfrac{1}{x+1}\end{array}\\\\\\\\ \displaystyle\int\!u\,dv=uv-\int\!v\,du\\\\\\ \int\!\frac{x\,e^x}{(x+1)^2}=x\,e^x\cdot \left(-\,\frac{1}{x+1} \right )-\int\!\left(-\,\frac{1}{x+1} \right )\cdot (x+1)\,e^x\,dx\\\\\\ I=-\,\frac{x\,e^x}{x+1}+\int\!\frac{1}{x+1}\cdot (x+1)\,e^x\,dx\\\\\\ I=-\,\frac{x\,e^x}{x+1}+\int\!e^x\,dx\\\\\\ I=-\,\frac{x\,e^x}{x+1}+e^x+C\\\\\\\\ \therefore~~\boxed{\begin{array}{c} \displaystyle\int\!\frac{x\cdot e^x}{(x+1)^2}\,dx=-\,\frac{x\,e^x}{x+1}+e^x+C \end{array}}$


Bons estudos! :-)

Answer 2

$\sf \displaystyle \int \frac{x\cdot e^x}{\left(x+1\right)^2}dx\\\\\\=-\frac{e^xx}{x+1}-\int \:-e^xdx\\\\\\=-\frac{e^xx}{x+1}-\left(-e^x\right)\\\\\\=-\frac{e^xx}{x+1}+e^x\\\\\\\to \boxed{\sf =-\frac{e^xx}{x+1}+e^x+C}$