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)

We wish to evaluate the integral $$\displaystyle \int \frac{dx}{(x+4)^{\frac{8}{7}}(x-3)^{\frac{6}{7}}}\,.$$

It looks convenient to create a fraction inside a single power, so we set $$u=\frac{x-3}{x+4}.$$

First we differentiate this substitution to express $$dx$$ in terms of $$du.$$ Using the quotient rule, we have

$$\frac{du}{dx}=\frac{(x+4)\cdot 1-(x-3)\cdot 1}{(x+4)^2}=\frac{x+4-x+3}{(x+4)^2}=\frac{7}{(x+4)^2}.$$

Hence $$du=\frac{7\,dx}{(x+4)^2}\quad\Longrightarrow\quad dx=\frac{(x+4)^2}{7}\,du.$$

Now we rewrite every factor of the integrand in terms of $$u$$ and $$(x+4).$$ From the definition of $$u,$$ we have $$(x-3)=u(x+4).$$ Therefore

$$\frac{1}{(x+4)^{\frac{8}{7}}(x-3)^{\frac{6}{7}}} =\frac{1}{(x+4)^{\frac{8}{7}}\big(u(x+4)\big)^{\frac{6}{7}}} =\frac{1}{(x+4)^{\frac{8}{7}}\,u^{\frac{6}{7}}(x+4)^{\frac{6}{7}}} =u^{-\frac{6}{7}}(x+4)^{-\frac{14}{7}} =u^{-\frac{6}{7}}(x+4)^{-2}.$$

Multiplying this by our expression for $$dx$$ gives

$$\begin{aligned} \frac{1}{(x+4)^{\frac{8}{7}}(x-3)^{\frac{6}{7}}}\,dx &=u^{-\frac{6}{7}}(x+4)^{-2}\;\frac{(x+4)^2}{7}\,du\\[4pt] &=\frac{1}{7}\,u^{-\frac{6}{7}}\,du. \end{aligned}$$

So the original integral converts neatly to

$$\int \frac{dx}{(x+4)^{\frac{8}{7}}(x-3)^{\frac{6}{7}}} =\frac{1}{7}\int u^{-\frac{6}{7}}\,du.$$

We now apply the power‐rule for integration, which states $$\displaystyle\int u^n\,du=\frac{u^{n+1}}{n+1}+C$$ provided $$n\neq -1.$$ Here $$n=-\frac{6}{7},$$ so $$n+1=\frac{1}{7}\neq-1.$$

Evaluating the integral, we get

$$\frac{1}{7}\int u^{-\frac{6}{7}}\,du =\frac{1}{7}\left[\frac{u^{-\frac{6}{7}+1}}{-\frac{6}{7}+1}\right]+C =\frac{1}{7}\left[\frac{u^{\frac{1}{7}}}{\frac{1}{7}}\right]+C =u^{\frac{1}{7}}+C.$$

Finally we substitute back $$u=\dfrac{x-3}{x+4}$$ to obtain

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

Hence, the correct answer is Option A.