Question

$\int \frac{dx}{{(x+\alpha )}^{8/7}{(x-\beta )}^{6/7}}=\frac{k}{\alpha +\beta }{\left(\frac{x-\beta }{x+\alpha }\right)}^{1/k}.$ Find the value of $k$.

Answer 1

Let $I=\int \frac{dx}{{(x+\alpha )}^{8/7}{(x-\beta )}^{6/7}}$
$I=\frac{dx}{{(x+\alpha )}^{8/7+6/7}\cdot {\left(\frac{x-\beta }{x+\alpha }\right)}^{8/7}}$
$=\frac{dx}{{(x+\alpha )}^2{[1-\frac{\alpha +\beta }{x+\alpha }]}^{6/7}}$
Substitute $1-\frac{\alpha +\beta }{x+\alpha }=t\therefore \frac{\alpha +\beta }{{(x+\alpha )}^2}dx=dt$
$\therefore I=\frac{1}{(\alpha +\beta )}\int \frac{dt}{t^{6/7}}=\frac{7}{\alpha +\beta }{t}^{1/7}$
$=\frac{7}{\alpha +\beta }{[1-\frac{\alpha +\beta }{x+\alpha }]}^{1/7}=\frac{7}{\alpha +\beta }{\left(\frac{x-\beta }{x+\alpha }\right)}^{1/7}$