What is the integral of x\sqrt[3]{(9+x^2)^2} ?

The integral of x\sqrt[3]{(9+x^2)^2} is 3/10 (9+x^2)^{5/3}+C

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integral of x\sqrt[3]{(9+x^2)^2}

\frac{3}{10} (9 + x^{2})^{\frac{5}{3}} + C

Solution steps

Apply u-substitution

= \int \frac{( 9 + u ) ^{\frac{2}{3}}}{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 (9 + u)^{\frac{2}{3}} d u

Apply u-substitution

= \frac{1}{2} \cdot \int v^{\frac{2}{3}} d v

Apply the Power Rule

= \frac{1}{2} \cdot \frac{3}{5} v^{\frac{5}{3}}

Substitute back

= \frac{1}{2} \cdot \frac{3}{5} (9 + x^{2})^{\frac{5}{3}}

Simplify

\frac{1}{2} \cdot \frac{3}{5} (9 + x^{2})^{\frac{5}{3}} : \frac{3}{10} (9 + x^{2})^{\frac{5}{3}}

= \frac{3}{10} (9 + x^{2})^{\frac{5}{3}}

Add a constant to the solution

= \frac{3}{10} (9 + x^{2})^{\frac{5}{3}} + C

What is the integral of x\sqrt[3]{(9+x^2)^2} ?
The integral of x\sqrt[3]{(9+x^2)^2} is 3/10 (9+x^2)^{5/3}+C