Question

Evaluate:
$\int \frac{dx}{{\left(x+3\right)}^{15/16}{\left(x-4\right)}^{17/16}}.$

Answer 1

$I=\int \frac{dx}{(x+3{)}^{15/16}(x-4{)}^{17/16}}=\int \frac{dx(x-4{)}^{15/16}}{(x+3{)}^{15/16}(x-4{)}^2}$

$=\int \left[\frac{dx}{(x-4{)}^2}\right]{\left(\frac{x-4}{x+3}\right)}^{15/16}$

Put $\frac{x+3}{x-4}=t$

$\Rightarrow \frac{1(x-4)-1(x+3)}{(x-4{)}^2}dx=dt$

$\Rightarrow -7\times \frac{1}{(x-4{)}^2}dx=dt$

$\Rightarrow \left[\frac{dx}{(x-4{)}^2}\right]=\frac{-1}{7}dt$

$\therefore I=\int \frac{-dt}{7}\times {\frac{1}{t}}^{15/16}=\frac{-7}{7}\int t^{-15/16}dt$

$=\frac{1}{7}\frac{t^{-15/16+1}}{(-15/16)+1}+c$

$=\frac{-1}{7}\times 16t^{1/16}+c$

$=\frac{-16}{7}{\left(\frac{x+3}{x-4}\right)}^{1/16}+c$

$\therefore \int \frac{dx}{(x+3{)}^{15/16}(x-4{)}^{17/16}}=\frac{-16}{7}{\left(\frac{x+3}{x-4}\right)}^{1/16}+c$