Q115 / 160 · AP EAMCET 2022 (07 Jul Shift 2) · Mathematics

$\int_{\alpha+1}^\alpha \frac{e^x(\alpha-x)}{(x-\alpha+1)^2} d x=$

  • A
    $2 e^\alpha+e$
  • B
    $\frac{2 e^{\alpha+2}}{e-2}$
  • C
    $e^\alpha \frac{(e+2)}{2}$
  • Correct answer
    D
    $e^\alpha\left(\frac{e-2}{2}\right)$

Solution

$\begin{aligned} I & =\int_{\alpha+1}^\alpha \frac{e^x(\alpha-x)}{(x-\alpha+1)^2} d x=\int_{\alpha+1}^\alpha \frac{e^x(\alpha-x-1+1)}{(x-\alpha+1)^2} d x \\ & =\int_{\alpha+1}^\alpha e^x\left[\frac{1}{(x-\alpha+1)^2}-\frac{1}{(x-\alpha+1)}\right] d x \\ & =\left[-\frac{e^x}{(x-\alpha+1)}\right]_{\alpha+1}^\alpha \\ & =-e^\alpha+\frac{e^{\alpha+1}}{2}=e^\alpha\left[\frac{e-2}{2}\right]\end{aligned}$

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

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