Question:
The integral $\int \frac{d x}{(x+4)^{8 / 7}(x-3)^{6 / 7}}$ is equal to:
(where $\mathrm{C}$ is a constant of integration)
Correct Option: 1
Solution:
$I=\int \frac{d x}{(x+4)^{8 / 7}(x-3)^{6 / 7}}$
$=\int\left(\frac{x-3}{x+4}\right)^{\frac{-6}{7}} \frac{1}{(x+4)^{2}} d x$
Let $\frac{x-3}{x+4}=t^{7}$,
Differentiate on both sides, we get
$\frac{7}{(x+4)^{2}} d x=7 t^{6} d t$
Hence, $I=\int t^{-6} t^{6} d t=t+C=\left(\frac{x-3}{x+4}\right)^{\frac{1}{7}}+C$
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