What is the integral of x^4(9-x^5)^{1/4} ?

The integral of x^4(9-x^5)^{1/4} is -4/25 (9-x^5)^{5/4}+C

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integral of x^4(9-x^5)^{1/4}

\int x^{4} (9 - x^{5})^{\frac{1}{4}} d x

- \frac{4}{25} (9 - x^{5})^{\frac{5}{4}} + C

Solution steps

\int x^{4} (9 - x^{5})^{\frac{1}{4}} d x

Apply u-substitution

= \int - \frac{u ^{\frac{1}{4}}}{5} d u

Take the constant out:

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

= - \frac{1}{5} \cdot \int u^{\frac{1}{4}} d u

Apply the Power Rule

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

Substitute back

u = 9 - x^{5}

= - \frac{1}{5} \cdot \frac{4}{5} (9 - x^{5})^{\frac{5}{4}}

Simplify

- \frac{1}{5} \cdot \frac{4}{5} (9 - x^{5})^{\frac{5}{4}} : - \frac{4}{25} (9 - x^{5})^{\frac{5}{4}}

= - \frac{4}{25} (9 - x^{5})^{\frac{5}{4}}

Add a constant to the solution

= - \frac{4}{25} (9 - x^{5})^{\frac{5}{4}} + C

What is the integral of x^4(9-x^5)^{1/4} ?
The integral of x^4(9-x^5)^{1/4} is -4/25 (9-x^5)^{5/4}+C