What is the integral of x/(\sqrt[5]{x^2-1)} ?

The integral of x/(\sqrt[5]{x^2-1)} is 5/8 (x^2-1)^{4/5}+C

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integral of x/(\sqrt[5]{x^2-1)}

\frac{5}{8} (x^{2} - 1)^{\frac{4}{5}} + 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{5}{4} u^{\frac{4}{5}}

= \frac{1}{2} \cdot \frac{5}{4} u^{\frac{4}{5}}

Substitute back

u = x^{2} - 1

= \frac{1}{2} \cdot \frac{5}{4} (x^{2} - 1)^{\frac{4}{5}}

Simplify

\frac{1}{2} \cdot \frac{5}{4} (x^{2} - 1)^{\frac{4}{5}} : \frac{5}{8} (x^{2} - 1)^{\frac{4}{5}}

= \frac{5}{8} (x^{2} - 1)^{\frac{4}{5}}

Add a constant to the solution

= \frac{5}{8} (x^{2} - 1)^{\frac{4}{5}} + C

y = 4 x^{2}, y = 6 x^{2}, 5 x + y = 6

(2 x - 3)^{2}

\frac{\partial}{\partial x} (x e^{- x} (A x^{2} + B x + C))

\int - \frac{ln ( x )}{x} d x

\frac{d y}{y - y ^{2}} = d x

What is the integral of x/(\sqrt[5]{x^2-1)} ?
The integral of x/(\sqrt[5]{x^2-1)} is 5/8 (x^2-1)^{4/5}+C