Question

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

Answer 1

Let $t=x+2\Rightarrow dt=dx$

$=\int \frac{{(t-2)}^2{e}^{t-2}}{t^2}dt$

$=\int \frac{(t^2-4t+4)e^{t-2}}{t^2}dt$

$=\int \frac{t^2{e}^{t-2}}{t^2}dt-4\int \frac{t{e}^{t-2}}{t^2}dt+4\int \frac{e^{t-2}}{t^2}dt$

$=\int e^{t-2}dt-4\int \frac{e^{t-2}}{t}dt+4\int \frac{e^{t-2}}{t^2}dt$

$=e^{t-2}-4\int \frac{e^{t-2}}{t}dt+4\int \frac{e^{t-2}}{t^2}dt$

Let $u=e^{t-2}\Rightarrow du=e^{t-2}dt$

$dv=\frac{dt}{t^2}\Rightarrow v=-\frac{1}{t}$

$=e^{t-2}-4\int \frac{e^{t-2}}{t}dt+4[-\frac{1}{t}{e}^{t-2}+\int \frac{e^{t-2}dt}{t}dt]+c$

$=e^{t-2}-4\int \frac{e^{t-2}}{t}dt-4\frac{1}{t}{e}^{t-2}+4\int \frac{e^{t-2}dt}{t}dt+c$

$=e^{t-2}-4\frac{1}{t}{e}^{t-2}+c$

$=(1-\frac{4}{t})e^{t-2}+c$

$=(1-\frac{4}{x+2})e^{x+2-2}+c$

$=\frac{x+2-4}{x+2}{e}^x+c$

$=\frac{x-2}{x+2}{e}^x+c$