$∫\frac{dx}{(x+a)^{\frac{8}{7}}(x-b)^{\frac{6}{7}}}$ is equal to

$(\frac{7}{a+b})(\frac{x-b}{x+a}^{\frac{1}{7}})+c$

Let $I=∫\frac{dx}{(x+a)^{\frac{8}{7}}(x-b)^{\frac{6}{7}}}=∫\frac{dx}{(x+a)^2(\frac{x-b}{x+a}^{\frac{6}{7}})}$.

If $\frac{x-b}{x+a}=p$, then $\frac{a+b}{(x+a)^2}dx=dp⇒I=\frac{1}{a+b}∫\frac{dp}{p^{\frac{6}{7}}}$

$=\frac{7}{a+b}(p^{\frac{1}{7}})+c=(\frac{7}{a+b})(\frac{x-b}{x+a}^{\frac{1}{7}})+c$

Hence (B) is the correct answer.