Mathematics - Definite Integration Question with Solution | TestHub

$\mathrm{I}=\underset{\alpha }{\overset{\alpha +1}{\int }}\frac{dx}{(x+\alpha )(x+\alpha +1)}=\underset{\alpha }{\overset{\alpha +1}{\int }}\left(\frac{1}{x+\alpha }–\frac{1}{x+\alpha +1 }\right)dx$
Using partial fractions
$={\left[\mathrm{l}\mathrm{n}\left|x+\alpha \right|-\mathrm{l}\mathrm{n}\left|x+\alpha +1\right|\right]}_{\alpha }^{\alpha +1}$
$={\left[\mathrm{l}\mathrm{n}\left|\frac{x+\alpha }{x+\alpha +1}\right| \right]}_{\alpha }^{\alpha +1}$
$=\mathrm{l}\mathrm{n}\left|\frac{2\mathrm{\alpha }+1}{2\mathrm{\alpha }+2}\right|-\mathrm{l}\mathrm{n}\left|\frac{2\mathrm{\alpha }}{2\mathrm{\alpha }+1}\right|$
$=\mathrm{l}\mathrm{n}\left|\frac{{\left(2\alpha +1\right)}^2}{{\left(2\alpha +1\right)}^2-1}\right|=\ln \frac{9}{8}$
$\Rightarrow {(2\alpha +1)}^2=9$
$\Rightarrow 2\alpha +1=\pm 3$
$\Rightarrow \alpha =1$or$-2$.