What is the integral of \sqrt[5]{x^4} ?

The integral of \sqrt[5]{x^4} is 5/9 x^{9/5}+C

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integral of \sqrt[5]{x^4}

\int 5 x^{4} d x

\frac{5}{9} x^{\frac{9}{5}} + C

Solution steps

\int 5 x^{4} d x

5 x^{4} = x^{\frac{4}{5}},

assuming

x \geq 0

= \int x^{\frac{4}{5}} d x

Apply the Power Rule

= \frac{5}{9} x^{\frac{9}{5}}

Add a constant to the solution

= \frac{5}{9} x^{\frac{9}{5}} + C

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

\frac{( t ^{3} + t )}{( t ^{4} - 2 )}

y^{^{'}} + e^{x} y = y^{2}

\frac{1 - 4 x}{1 + 4 x}

y = - 3 x - 5

What is the integral of \sqrt[5]{x^4} ?
The integral of \sqrt[5]{x^4} is 5/9 x^{9/5}+C