What is the integral of x/(\sqrt[17]{x^2-1)} ?

The integral of x/(\sqrt[17]{x^2-1)} is 17/32 (x^2-1)^{16/17}+C

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integral of x/(\sqrt[17]{x^2-1)}

\frac{17}{32} (x^{2} - 1)^{\frac{16}{17}} + C

Solution steps

Apply u-substitution

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

Apply the Power Rule

: \frac{17}{16} u^{\frac{16}{17}}

= \frac{1}{2} \cdot \frac{17}{16} u^{\frac{16}{17}}

Substitute back

u = x^{2} - 1

= \frac{1}{2} \cdot \frac{17}{16} (x^{2} - 1)^{\frac{16}{17}}

Simplify

\frac{1}{2} \cdot \frac{17}{16} (x^{2} - 1)^{\frac{16}{17}} : \frac{17}{32} (x^{2} - 1)^{\frac{16}{17}}

= \frac{17}{32} (x^{2} - 1)^{\frac{16}{17}}

Add a constant to the solution

= \frac{17}{32} (x^{2} - 1)^{\frac{16}{17}} + C

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What is the integral of x/(\sqrt[17]{x^2-1)} ?
The integral of x/(\sqrt[17]{x^2-1)} is 17/32 (x^2-1)^{16/17}+C