Question

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

Answer 1

$\int \frac{x^2{e}^x}{{(x+2)}^2}$

$=\int x^2{e}^x\frac{1}{{(x+2)}^2}dx$ [ Using integration by partly ]

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

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

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

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

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

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

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

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

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

Hence, the answer is $\frac{{-x}^2{e}^x}{(x+2)}+x{e}^x-e^x+c.$