Question

$\int e^x\frac{x-1}{(x+1{)}^3}dx$ is equal to

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

Answer 1

Answer: (B)

$\frac{e^x}{(x+1{)}^2}+C$

$\int e^x\frac{x-1}{(x+1{)}^3}dx=\int e^x\left(\frac{(x+1)-2}{(x+1{)}^3}\right)dx$
$=\int e^x(\frac{1}{(x+1{)}^2}+\frac{-2}{(x+1{)}^3})dx$

$=\frac{e^x}{(x+1{)}^2}+C$
Since, $\int e^x\left[f\right(x)+f^{\prime }(x\left)\right]dx=e^{x}f\left(x\right)+c$