int_{alpha+1}^{alpha} frac{e^x(alpha-x)}{(x-a | vedclass

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(D) Let $I = \int_{\alpha+1}^{\alpha} \frac{e^x(\alpha-x)}{(x-\alpha+1)^2} dx$. Substitute $u = x - \alpha + 1$,then $du = dx$. When $x = \alpha+1$,$u

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $I = \int_{\alpha+1}^{\alpha} \frac{e^x(\alpha-x)}{(x-\alpha+1)^2} dx$. Substitute $u = x - \alpha + 1$,then $du = dx$. When $x = \alpha+1$,$u = 2$. When $x = \alpha$,$u = 1$. Also,$\alpha - x = 1 - u$. So,$I = \int_{2}^{1} \frac{e^{u+\alpha-1}(1-u)}{u^2} du = e^{\alpha-1} \int_{2}^{1} e^u \left(\frac{1}{u^2} - \frac{1}{u}\right) du$. Using the formula $\int e^u (f(u) + f'(u)) du = e^u f(u) + C$,where $f(u) = -\frac{1}{u}$ and $f'(u) = \frac{1}{u^2}$. $I = e^{\alpha-1} \left[ e^u \left(-\frac{1}{u}\right) \right]_{2}^{1} = e^{\alpha-1} \left[ -e^1 + \frac{e^2}{2} \right] = e^{\alpha-1} \left[ \frac{e^2 - 2e}{2} \right] = e^{\alpha} \left( \frac{e-2}{2} \right)$.

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

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