Q65 / 180 · KCET 2013 · Mathematics

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

  • A
    $\frac{e^{x}}{x+1}+C$
  • Correct answer
    B
    $\frac{e^{x}}{(x+1)^{2}}+C$
  • C
    $\frac{e^{x}}{(x+1)^{3}}+C$
  • D
    $\frac{x \cdot e^{x}}{(x+1)}+C$

Solution

Let $I=\int \frac{x-1}{(x+1)^{3}} e^{x} d x$ $=\int \frac{(x+1)-2}{(x+1)^{3}} \cdot e^{x} d x$ $=\int \frac{e^{x}}{(x+1)^{2}} d x-2 \int \frac{e^{x}}{(x+1)^{3}} d x$ $=\frac{1}{(x+1)^{2}} \cdot e^{x}-\int \frac{-2}{(1+x)^{3}} \cdot e^{x} d x-2 \int \frac{e^{x}}{(x+1)^{3}} d x$ $=\frac{e^{x}}{(x+1)^{2}}+C$

Hence, the correct answer is option: 'B'.

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