Question

$\int \frac{x^2{e}^x}{{(x+2)}^2}dx$ is equal to

• A. $e^x\left[\frac{x-2}{x+2}\right]+C$
• B. $e^x\left[\frac{4}{{(x+2)}^2}\right]+C$
• C. $e^x\left[\frac{1}{x+2}\right]+C$
• D. $e^x\left[\frac{x^2}{{(x+2)}^2}\right]+C$

Answer 1

The correct option is A $e^x\left[\frac{x-2}{x+2}\right]+C$

$I=\int \frac{x^2{e}^x}{{(x+2)}^2}dx=\int e^x\frac{x^2}{{(x+2)}^2}dx=\int e^x\frac{x^2-4+4}{{(x+2)}^2}dx=\int e^x\{\frac{x^2-4}{{(x+2)}^2}+\frac{4}{{(x+2)}^2}\}dx=\int e^x\{\frac{x-2}{x+2}+\frac{4}{{(x+2)}^2}\}dx\underset{f\left(x\right)}{\downarrow }\underset{f^{\prime }\left(x\right)}{\downarrow }\{\because \int e^x\left\{f\right(x)+f^{\prime }(x\left)\right\}dx=e^{x}f\left(x\right)+C\}=e^x\left(\frac{x-2}{x+2}\right)+C$