Step 1: Observe the structure of the integrand.
We need $\int \frac{dx}{(x-3)^{4/5}(x+1)^{6/5}}$. Note the exponents: $4/5 + 6/5 = 2$. This suggests the substitution $t = \frac{x-3}{x+1}$, which involves the ratio of the two linear factors.
Step 2: Compute $dt/dx$.
$t = \frac{x-3}{x+1}$, so $\frac{dt}{dx} = \frac{(x+1) - (x-3)}{(x+1)^2} = \frac{4}{(x+1)^2}$. Hence $dx = \frac{(x+1)^2}{4}\,dt$.
Step 3: Express the denominator in terms of $t$.
$(x-3)^{4/5}(x+1)^{6/5} = (x+1)^{4/5} \cdot \left(\frac{x-3}{x+1}\right)^{4/5} \cdot (x+1)^{6/5}$. Wait, let us factor directly: $(x-3)^{4/5}(x+1)^{6/5} = (x+1)^{4/5+6/5}\cdot\left(\frac{x-3}{x+1}\right)^{4/5} = (x+1)^2 \cdot t^{4/5}$.
Step 4: Substitute into the integral.
$\int \frac{(x+1)^2/4}{(x+1)^2 \cdot t^{4/5}}\,dt = \frac{1}{4}\int t^{-4/5}\,dt = \frac{1}{4} \cdot \frac{t^{1/5}}{1/5} + C = \frac{5}{4}\,t^{1/5} + C$.
Step 5: Substitute $t$ back.
$= \frac{5}{4}\left(\frac{x-3}{x+1}\right)^{1/5} + C$.
Step 6: State the final answer.
\[\boxed{\dfrac{5}{4}\left(\dfrac{x-3}{x+1}\right)^{1/5} + C}\]