What is the integral of (x^2+x^5)\sqrt[3]{x^6+2x^3} ?

The integral of (x^2+x^5)\sqrt[3]{x^6+2x^3} is 1/8 x^4(x^3+2)^{4/3}+C

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integral of (x^2+x^5)\sqrt[3]{x^6+2x^3}

\frac{1}{8} x^{4} (x^{3} + 2)^{\frac{4}{3}} + 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{1}{6} \cdot \frac{3}{4} u^{\frac{4}{3}}

Substitute back

u = x^{6} + 2 x^{3}

= \frac{1}{6} \cdot \frac{3}{4} (x^{6} + 2 x^{3})^{\frac{4}{3}}

Simplify

\frac{1}{6} \cdot \frac{3}{4} (x^{6} + 2 x^{3})^{\frac{4}{3}} : \frac{1}{8} x^{4} (x^{3} + 2)^{\frac{4}{3}}

= \frac{1}{8} x^{4} (x^{3} + 2)^{\frac{4}{3}}

Add a constant to the solution

= \frac{1}{8} x^{4} (x^{3} + 2)^{\frac{4}{3}} + C

What is the integral of (x^2+x^5)\sqrt[3]{x^6+2x^3} ?
The integral of (x^2+x^5)\sqrt[3]{x^6+2x^3} is 1/8 x^4(x^3+2)^{4/3}+C