Question

A value of $\alpha$ such that $\int_{\alpha}^{\alpha + 1} \frac{dx}{(x + \alpha) (x + \alpha + 1} = log_e (\frac{9}{8})$is :

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

Answer 1

Answer: (D)

$\int_{\alpha}^{\alpha + 1} \, \frac{(x +\alpha + 1 )-(x + \alpha)}{(x+ \alpha)(x + \alpha + 1)}$dx = $(\ell n |x+\alpha| - \ell n |x + \alpha + 1|)_n^{\alpha + 1}$ $= \ell n |\frac{2\alpha + 1}{2\alpha + 2} \times \frac{2\alpha + 1}{2\alpha}| = \ell n \frac{9}{8}$
$\Rightarrow \, \, = -2, 1$