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AP EAMCET · Maths · Definite Integration

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

  1. A \(2 e^\alpha+e\)
  2. B \(\frac{2 e^{\alpha+2}}{e-2}\)
  3. C \(e^\alpha \frac{(e+2)}{2}\)
  4. D \(e^\alpha\left(\frac{e-2}{2}\right)\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(e^\alpha\left(\frac{e-2}{2}\right)\)

Step-by-step Solution

Detailed explanation

\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 \\ &…