$\int \frac{x-3}{(x-1)^3} \mathrm{e}^x \mathrm{~d} x=$

$\int \frac{x-3}{(x-1)^3} \mathrm{e}^x \mathrm{~d} x=$
  1. $\mathrm{e}^x\left(\frac{1}{(x-1)^2}\right)+\mathrm{c}$, where $\mathrm{c}$ is constant of integration,
  2. $\mathrm{e}^x\left(\frac{1}{x+1}\right)+\mathrm{c}$, where $\mathrm{c}$ is constant of integration.
  3. $\mathrm{e}^x\left((x-1)^2\right)+\mathrm{c}$, where $\mathrm{c}$ is constant of integration.
  4. $\mathrm{e}^x\left((x-1)^3\right)+\mathrm{c}$, where $\mathrm{c}$ is constant of integration.

Solution

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

Asked in: MHT CET 2023 (13 May Shift 2)

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