Question

Let $\int \frac{dx}{(x+a{)}^{\frac{8}{7}}(x-b{)}^{\frac{6}{7}}}=\mu (x-b{)}^{\frac{1}{7}}+C$, where ${}^{\prime }{\mu }^{\prime }$; is a function of ${}^{\prime }{x}^{\prime }$, then $\left[\frac{\mu (a+b)}{(x+a{)}^{\frac{1}{7}}}\right]$ equals to

Answer 1

Substitue
$\frac{x-b}{(x+a)}=t\Rightarrow \frac{(a+b)}{(x+a{)}^2}dx=dt$

Given Integration reduces to
$I=\frac{1}{(a+b)}\int t^{\frac{6}{7}}.dt$
$\Rightarrow I=\frac{7}{(a+b)}.\frac{(x-b{)}^{\frac{1}{7}}}{(x+a{)}^{\frac{1}{7}}}+C$

According to Question
$\mu =\frac{7}{(a+b)(x+a{)}^{\frac{1}{7}}}\Rightarrow \frac{\mu (a+b)}{(x+a{)}^{-\frac{1}{7}}}=7$