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