The value of $\mathrm{I}=\int \frac{(x-1) \mathrm{e}^x}{(x+1)^3} \mathrm{~d} x$ is
The value of $\mathrm{I}=\int \frac{(x-1) \mathrm{e}^x}{(x+1)^3} \mathrm{~d} x$ is
- $\frac{-\mathrm{e}^x}{(x+1)^2}+\mathrm{C}$, (where C is a constant of
integration)
- $\frac{-x \mathrm{e}^x}{(x+1)^2}+\mathrm{C}$, (where C is a constant of integration)
- $\frac{x \mathrm{e}^x}{(x+1)^2}+\mathrm{C}$, (where C is a constant of integration)
- $\frac{\mathrm{e}^x}{(x+1)^2}+\mathrm{C}$, (where C is a constant of integration)
Solution
$\begin{aligned} I & =\int \frac{(x-1) \mathrm{e}^x}{(x+1)^3} \mathrm{~d} x \\ I & =\int\left(\frac{x+1-2}{(x+1)^3}\right) \mathrm{e}^x \mathrm{~d} x \\ & =\int\left[\frac{x+1}{(x+1)^3}-\frac{2}{(x+1)^3}\right] \mathrm{e}^x \mathrm{~d} x \\ & =\int\left[\frac{1}{(x+1)^2}-\frac{2}{(x+1)^3}\right] \mathrm{e}^x \mathrm{~d} x \\ & =\mathrm{e}^x\left(\frac{1}{(x+1)^2}\right)+\mathrm{c}\end{aligned}$
$\ldots\left\{\int \mathrm{e}^x\left(\mathrm{f}(x)+\mathrm{f}^{\prime}(x)\right) \mathrm{d} x=\mathrm{e}^x \mathrm{f}(x)+\mathrm{c}\right\}$
Asked in: MHT CET 2024 (09 May Shift 1)
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