Question

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 :

• A. 2
• B. - 2
• C. \(\frac{1}{2}\)
• D. \(-\frac{1}{2}\)

Hint:

Use partial fractions \(\frac{1}{(x+\alpha)(x+\alpha+1)} = \frac{1}{x+\alpha}-\frac{1}{x+\alpha+1}\) and solve the resulting quadratic in \(\alpha\).

Answer 1

The Correct Option is B

\(\frac{1}{(x+\alpha)(x+\alpha+1)} = \frac{1}{x+\alpha} - \frac{1}{x+\alpha+1}\).
\(I = \left[\log_e\left|\frac{x+\alpha}{x+\alpha+1}\right|\right]_{\alpha}^{\alpha+1} = \log_e\frac{2\alpha+1}{2\alpha+2} - \log_e\frac{2\alpha}{2\alpha+1}\).
\(= \log_e\frac{(2\alpha+1)^2}{4\alpha(\alpha+1)}\). Set equal to \(\log_e\frac{9}{8}\): \(8(2\alpha+1)^2 = 9\cdot 4\alpha(\alpha+1)\).
\(32\alpha^2 + 32\alpha + 8 = 36\alpha^2 + 36\alpha \Rightarrow \alpha^2 + \alpha - 2 = 0 \Rightarrow \alpha = 1\) or \(\alpha = -2\).
Among the options only \(\alpha = -2\) is present.
Answer: (2) - 2