Question

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

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

Answer 1

Answer: (A)

$\left(\frac{x-3}{x+4}\right)^{\frac{1}{7}}+C$

Answer 2

The given integral is solved using the substitution $t = \frac{x-3}{x+4}$. After differentiating and substituting, the integral becomes $\int t^{-\frac{6}{7}} dt$, which evaluates to $t^{\frac{1}{7}} + C$. Substituting back gives the final answer $\left(\frac{x-3}{x+4}\right)^{\frac{1}{7}}+C$.