Question

Evaluate $\int \frac{(x-1) e^{x}}{(x+1)^{3}} d x$

• A. $ 0 $
• B. $\frac{(x+1)^{2}}{e^{x}}+C$
• C. $\frac{e^{x}}{(x+1)^{2}}+C$
• D. $\frac{e^{x}}{(x-1)^{2}}+(x+1)^{2}+C$

Answer 1

Answer: (C)

Let $I=\int \frac{(x-1) e^{x}}{(x+1)^{3}} d x$
$\Rightarrow I=\int\left\{\frac{x+1-2}{(x+1)^{3}}\right\} e^{x} d x$
$=\int\left\{\frac{1}{(x+1)^{2}}-\frac{2}{(x+1)^{3}}\right\} e^{x} d x$
$\int e^{x} \cdot \frac{1}{(x+1)^{2}} d x-2 \int e^{x} \frac{1}{(x+1)^{3}} d x$
Applying integrating by parts, we get
$=\left(\frac{1}{(x+1)^{2}} e^{x}-\int e^{x} \frac{(-2)}{(x+1)^{3}} d x\right)$
$-2 \int e^{x} \frac{1}{(x+1)^{3}} d x$
$=\frac{e^{x}}{(x+1)^{2}}+C$
Note We can use the formula
$\int e^{x}\left[f(x)+f^{\prime}(x)\right] d x=e^{x} f(x)+C$