The integral


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)

  1. (1) $\left(\frac{x-3}{x+4}\right)^{1 / 7}+\mathrm{C}$

  2. (2) $-\left(\frac{x-3}{x+4}\right)^{-1 / 7}+\mathrm{C}$

  3. (3) $\frac{1}{2}\left(\frac{x-3}{x+4}\right)^{3 / 7}+\mathrm{C}$

  4. (4) $-\frac{1}{13}\left(\frac{x-3}{x+4}\right)^{-13 / 7}+\mathrm{C}$

Correct Option: 1


$I=\int \frac{d x}{(x+4)^{8 / 7}(x-3)^{6 / 7}}$

$=\int\left(\frac{x-3}{x+4}\right)^{\frac{-6}{7}} \frac{1}{(x+4)^{2}} d x$

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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