Question

$\int \frac{x{e}^x}{{(1+x)}^2}dx$ is equal to
(where $c$ is constant of integration)

• A. $\frac{e^x}{1+x^2}+c$
• B. $\frac{e^x}{{(1+x)}^2}+c$
• C. $\frac{e^x}{x}+c$
• D. $\frac{e^x}{1+x}+c$

Answer 1

The correct option is D $\frac{e^x}{1+x}+c$

$I=\int \frac{x{e}^x}{{(1+x)}^2}dx=\int e^x\frac{1+x-1}{{(1+x)}^2}dx=\int e^x[\frac{1}{1+x}+\frac{-1}{{(1+x)}^2}]dx\underset{f\left(x\right)}{\downarrow }\underset{f^{\prime }\left(x\right)}{\downarrow }=\frac{e^x}{1+x}+c[\because \int e^x\left\{f\right(x)+f^{\prime }(x\left)\right\}dx=e^{x}f\left(x\right)+c]$