The value of the integral $\int \frac{1}{(x-3)^{6 / 7}(x+4)^{8 / 7}} d x$ is

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

$7\left(\frac{x-3}{x+4}\right)^{6 / 7}+C$

$7\left(\frac{x+4}{x-3}\right)^{6 / 7}+C$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

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Analyzing +4 | -1

The value of the integral

\int \frac{1}{(x-3)^{6 / 7}(x+4)^{8 / 7}} d x

View answer and explanation

Correct option: A

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

Option A is correct because after rewriting the integral and using a simple substitution, the integral evaluates to

(\frac{x-3}{x+4})^{1 / 7}+C