1. We need to evaluate $I=\int \frac{dx}{(x+4)^{8/7}(x-3)^{6/7}}.$

2. Observe the expression $\frac{x-3}{x+4}$ because its derivative simplifies nicely: $\frac{d}{dx}\left(\frac{x-3}{x+4}\right)=\frac{(x+4)-(x-3)}{(x+4)^2}=\frac{7}{(x+4)^2}.$

3. Let $t=\left(\frac{x-3}{x+4}\right)^{1/7}.$ Then $t^7=\frac{x-3}{x+4}.$ Differentiate both sides: $7t^6\frac{dt}{dx}=\frac{7}{(x+4)^2}.$ So, $t^6\frac{dt}{dx}=\frac{1}{(x+4)^2}.$ Hence, $\frac{dt}{dx}=\frac{1}{t^6 (x+4)^2}.$

4. Now express $t^6$ in terms of $x$: $t^6=\left(\frac{x-3}{x+4}\right)^{6/7}=\frac{(x-3)^{6/7}}{(x+4)^{6/7}}.$ Therefore, $t^6(x+4)^2=\frac{(x-3)^{6/7}}{(x+4)^{6/7}}(x+4)^2=(x-3)^{6/7}(x+4)^{8/7}.$ So, $\frac{dt}{dx}=\frac{1}{(x+4)^{8/7}(x-3)^{6/7}}.$

5. This is exactly the integrand. Hence, $I=\int \frac{dx}{(x+4)^{8/7}(x-3)^{6/7}}=\int dt=t+C.$ Substituting back, $I=\left(\frac{x-3}{x+4}\right)^{1/7}+C.$

6. Compare with the options:

• A: incorrect power and coefficient
• B: matches exactly
• C: incorrect
• D: incorrect

Therefore, the correct option is B.