$\int \frac{\mathrm{d} x}{(x+a)^{\frac{9}{7}}(x-b)^{5 / 7}}=$

Evaluate the integral $I = \int \frac{\mathrm{d} x}{(x+a)^{\frac{9}{7}}(x-b)^{5 / 7}}$ using the substitution $t = \frac{x-b}{x+a}$.

Solving for $x$ in terms of $t$ gives $x = \frac{b+ta}{1-t}$, from which we obtain $\mathrm{d}x = \frac{a+b}{(1-t)^2} \mathrm{d}t$. The expressions transform as $x+a = \frac{a+b}{1-t}$ and $x-b = \frac{t(a+b)}{1-t}$.

Substituting into the integral yields:

$I = \int \frac{\frac{a+b}{(1-t)^2} \mathrm{d}t}{(\frac{a+b}{1-t})^{\frac{9}{7}} (\frac{t(a+b)}{1-t})^{\frac{5}{7}}}$

Simplifying the powers:

$I = \frac{1}{a+b} \int t^{-\frac{5}{7}} \mathrm{d}t = \frac{1}{a+b} \cdot \frac{t^{\frac{2}{7}}}{\frac{2}{7}} + C = \frac{7}{2(a+b)} t^{\frac{2}{7}} + C$

Substituting back $t = \frac{x-b}{x+a}$ gives the final result:

$I = \frac{7}{2(a+b)} \left(\frac{x-b}{x+a}\right)^{\frac{2}{7}} + C$

This corresponds to option C.