AP EAMCET · Maths · Definite Integration
\(\int_{\alpha+1}^\alpha \frac{e^x(\alpha-x)}{(x-\alpha+1)^2} d x=\)
- A \(2 e^\alpha+e\)
- B \(\frac{2 e^{\alpha+2}}{e-2}\)
- C \(e^\alpha \frac{(e+2)}{2}\)
- D \(e^\alpha\left(\frac{e-2}{2}\right)\)
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 \\ &…
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