Question

The integral $\int\frac{dx}{\left(x+4\right)^{8/7}\left(x-3\right)^{6/7}}$ is equal to :
(where C is a constant of integration)

• A. $-\left(\frac{x-3}{x+4}\right)^{-1/7} +C$
• B. $\frac{1}{2}\left(\frac{x-3}{x+4}\right)^{3/7} +C$
• C. $\left(\frac{x-3}{x+4}\right)^{1/7} +C$
• D. $-\frac{1}{13}\left(\frac{x-3}{x+4}\right)^{-13/7} +C$

Answer 1

Answer: (C)

$I=\int\frac{dx}{\left(x+4\right)^{\frac{8}{7}}\left(x-3\right)^{\frac{6}{7}}}=\int \frac{dx}{\left(\frac{x+4}{x-3}\right)^{\frac{8}{7}}\left(x-3\right)^{2}}$
Let $\frac{x+4}{x-3}=t \Rightarrow \frac{dx}{\left(x-3\right)^{2}}=-\frac{1}{7}dt$
$\Rightarrow I=-\frac{1}{7}\int \frac{dt}{t^{8/7}}=-\frac{1}{7}\int t^{-8/7}dt$
$=t^{-1/7}+C=+\left(\frac{x+4}{x-3}\right)^{-1/7}+C=\left(\frac{x-3}{x+4}\right)^{1/7}+C$