MHT CET · Maths · Indefinite Integration
The value of \(\mathrm{I}=\int \frac{(x-1) \mathrm{e}^x}{(x+1)^3} \mathrm{~d} x\) is
- A \(\frac{-\mathrm{e}^x}{(x+1)^2}+\mathrm{C}\), (where C is a constant of integration)
- B \(\frac{-x \mathrm{e}^x}{(x+1)^2}+\mathrm{C}\), (where C is a constant of integration)
- C \(\frac{x \mathrm{e}^x}{(x+1)^2}+\mathrm{C}\), (where C is a constant of integration)
- D \(\frac{\mathrm{e}^x}{(x+1)^2}+\mathrm{C}\), (where C is a constant of integration)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{e}^x}{(x+1)^2}+\mathrm{C}\), (where C is a constant of integration)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} I & =\int \frac{(x-1) \mathrm{e}^x}{(x+1)^3} \mathrm{~d} x \\ I & =\int\left(\frac{x+1-2}{(x+1)^3}\right) \mathrm{e}^x \mathrm{~d} x \\ & =\int\left[\frac{x+1}{(x+1)^3}-\frac{2}{(x+1)^3}\right] \mathrm{e}^x \mathrm{~d} x \\ & =\int\left[\frac{1}{(x+1)^2}-\frac{2}{(x+1)^3}\right] \mathrm{e}^x \mathrm{~d} x \\ & =\mathrm{e}^x\left(\frac{1}{(x+1)^2}\right)+\mathrm{c}\end{aligned}\)
\(\ldots\left\{\int \mathrm{e}^x\left(\mathrm{f}(x)+\mathrm{f}^{\prime}(x)\right) \mathrm{d} x=\mathrm{e}^x \mathrm{f}(x)+\mathrm{c}\right\}\)
\(\ldots\left\{\int \mathrm{e}^x\left(\mathrm{f}(x)+\mathrm{f}^{\prime}(x)\right) \mathrm{d} x=\mathrm{e}^x \mathrm{f}(x)+\mathrm{c}\right\}\)
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