What is the integral of (x^3-1)/(\sqrt[5]{x^4-4x)} ?

The integral of (x^3-1)/(\sqrt[5]{x^4-4x)} is 5/16 (x^4-4x)^{4/5}+C

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integral of (x^3-1)/(\sqrt[5]{x^4-4x)}

\frac{5}{16} (x^{4} - 4 x)^{\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}{4} \cdot \frac{5}{4} u^{\frac{4}{5}}

Substitute back

u = x^{4} - 4 x

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

Simplify

\frac{1}{4} \cdot \frac{5}{4} (x^{4} - 4 x)^{\frac{4}{5}} : \frac{5}{16} (x^{4} - 4 x)^{\frac{4}{5}}

= \frac{5}{16} (x^{4} - 4 x)^{\frac{4}{5}}

Add a constant to the solution

= \frac{5}{16} (x^{4} - 4 x)^{\frac{4}{5}} + C

(- 5, - 2), (5, 12)

y^{^{''}} + \frac{1}{2} y^{^{'}} + \frac{1}{16} y = 0, y (0) = 0, y^{^{'}} (0) = 5

\frac{\partial}{\partial y} (e^{x y} - z)

y^{^{''}} + 12 y^{^{'}} + 37 y = 0

x \to - 5 - lim (\frac{- 2}{x ^{2} - 25})

What is the integral of (x^3-1)/(\sqrt[5]{x^4-4x)} ?
The integral of (x^3-1)/(\sqrt[5]{x^4-4x)} is 5/16 (x^4-4x)^{4/5}+C