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MHT CET · Maths · Indefinite Integration

The value of \(\int \frac{(x-1) \mathrm{e}^x}{(x+1)^3} \mathrm{~d} x\) is equal to

  1. A \(\frac{\mathrm{e}^x}{(x+1)}+\mathrm{c}\), (where c is constant of integration)
  2. B \(\frac{\mathrm{e}^x}{(x+1)^2}+\mathrm{c}\), (where c is constant of integration)
  3. C \(\frac{-\mathrm{e}^x}{(x+1)}+\mathrm{c}\), (where c is constant of integration)
  4. D \(\frac{-\mathrm{e}^x}{(x+1)^2}+\mathrm{c}\), (where c is constant of ' integration)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{\mathrm{e}^x}{(x+1)^2}+\mathrm{c}\), (where c is constant of integration)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \text { Let } \mathrm{I}=\int \frac{(x-1) \mathrm{e}^x}{(x+1)^3} \mathrm{~d} x \\ & \begin{array}{l}=\int \frac{(x+1-2) \mathrm{e}^x}{(x+1)^3} \mathrm{~d} x \\ =\int \mathrm{e}^x\left[\frac{1}{(x+1)^2}-\frac{2}{(x+1)^3}\right] \mathrm{d} x \\ =\frac{\mathrm{e}^x}{(x+1)^2}+\mathrm{c} \\ \quad \ldots\left[\because \int \mathrm{e}^x\left[\mathrm{f}(x)+\mathrm{f}^{\prime}(x)\right] \mathrm{d} x=\mathrm{e}^x \mathrm{f}(x)+\mathrm{c}\right]\end{array}\end{aligned}\)