What is the integral of x^3(9-x^2)^{1/5} ?

The integral of x^3(9-x^2)^{1/5} is -1/2 (-5/11 (9-x^2)^{11/5}+15/2 (9-x^2)^{6/5})+C

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integral of x^3(9-x^2)^{1/5}

\int x^{3} (9 - x^{2})^{\frac{1}{5}} d x

- \frac{1}{2} (- \frac{5}{11} (9 - x^{2})^{\frac{11}{5}} + \frac{15}{2} (9 - x^{2})^{\frac{6}{5}}) + C

Solution steps

\int x^{3} (9 - x^{2})^{\frac{1}{5}} d x

Apply u-substitution

= \int - \frac{( - u + 9 ) u ^{\frac{1}{5}}}{2} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int (- u + 9) u^{\frac{1}{5}} d u

Expand

(- u + 9) u^{\frac{1}{5}} : - u^{\frac{6}{5}} + 9 u^{\frac{1}{5}}

= - \frac{1}{2} \cdot \int - u^{\frac{6}{5}} + 9 u^{\frac{1}{5}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \frac{1}{2} (- \int u^{\frac{6}{5}} d u + \int 9 u^{\frac{1}{5}} d u)

\int u^{\frac{6}{5}} d u = \frac{5}{11} u^{\frac{11}{5}}

\int 9 u^{\frac{1}{5}} d u = \frac{15}{2} u^{\frac{6}{5}}

= - \frac{1}{2} (- \frac{5}{11} u^{\frac{11}{5}} + \frac{15}{2} u^{\frac{6}{5}})

Substitute back

u = 9 - x^{2}

= - \frac{1}{2} (- \frac{5}{11} (9 - x^{2})^{\frac{11}{5}} + \frac{15}{2} (9 - x^{2})^{\frac{6}{5}})

Add a constant to the solution

= - \frac{1}{2} (- \frac{5}{11} (9 - x^{2})^{\frac{11}{5}} + \frac{15}{2} (9 - x^{2})^{\frac{6}{5}}) + C

What is the integral of x^3(9-x^2)^{1/5} ?
The integral of x^3(9-x^2)^{1/5} is -1/2 (-5/11 (9-x^2)^{11/5}+15/2 (9-x^2)^{6/5})+C