A value of alpha such that int_{alpha}^{alpha | vedclass

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(B) Given the integral $\int_{\alpha}^{\alpha+1} \frac{dx}{(x+\alpha)(x+\alpha+1)} = \log_{e}\left(\frac{9}{8}\right)$.Using partial fractions,$\frac{

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(B) Given the integral $\int_{\alpha}^{\alpha+1} \frac{dx}{(x+\alpha)(x+\alpha+1)} = \log_{e}\left(\frac{9}{8}\right)$.Using partial fractions,$\frac{1}{(x+\alpha)(x+\alpha+1)} = \frac{1}{x+\alpha} - \frac{1}{x+\alpha+1}$.Integrating both terms:$\int_{\alpha}^{\alpha+1} \left( \frac{1}{x+\alpha} - \frac{1}{x+\alpha+1} \right) dx = \left[ \log_{e}|x+\alpha| - \log_{e}|x+\alpha+1| \right]_{\alpha}^{\alpha+1} = \left[ \log_{e}\left| \frac{x+\alpha}{x+\alpha+1} \right| \right]_{\alpha}^{\alpha+1}$.Evaluating at the limits:$\log_{e}\left( \frac{2\alpha+1}{2\alpha+2} \right) - \log_{e}\left( \frac{2\alpha}{2\alpha+1} \right) = \log_{e}\left( \frac{(2\alpha+1)^2}{2\alpha(2\alpha+2)} \right) = \log_{e}\left( \frac{9}{8} \right)$.Equating the arguments:$\frac{(2\alpha+1)^2}{4\alpha(\alpha+1)} = \frac{9}{8} \Rightarrow 8(4\alpha^2 + 4\alpha + 1) = 9(4\alpha^2 + 4\alpha)$.$32\alpha^2 + 32\alpha + 8 = 36\alpha^2 + 36\alpha \Rightarrow 4\alpha^2 + 4\alpha - 8 = 0$.$\alpha^2 + \alpha - 2 = 0 \Rightarrow (\alpha+2)(\alpha-1) = 0$.Thus,$\alpha = 1$ or $\alpha = -2$. Comparing with the options,the correct answer is $-2$.

$A$ value of $\alpha$ such that $\int_{\alpha}^{\alpha+1} \frac{dx}{(x+\alpha)(x+\alpha+1)} = \log_{e}\left(\frac{9}{8}\right)$ is

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