Question

A value of $\alpha$ such that ${\int }_{\alpha }^{\alpha +1}\frac{dx}{(x+\alpha )(x+\alpha +1}=lo{g}_e(\frac{9}{8})$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 = $(\ell n|x+\alpha |-\ell n|x+\alpha +1|{)}_n^{\alpha +1}$ $=\ell n|\frac{2\alpha +1}{2\alpha +2}\times \frac{2\alpha +1}{2\alpha }|=\ell n\frac{9}{8}$
$\Rightarrow =-2,1$