What is the integral of \sqrt[3]{1+2x^2}x^3 ?

The integral of \sqrt[3]{1+2x^2}x^3 is 1/8 (3/7 (1+2x^2)^{7/3}-3/4 (1+2x^2)^{4/3})+C

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integral of \sqrt[3]{1+2x^2}x^3

\int 3 1 + 2 x^{2} x^{3} d x

\frac{1}{8} (\frac{3}{7} (1 + 2 x^{2})^{\frac{7}{3}} - \frac{3}{4} (1 + 2 x^{2})^{\frac{4}{3}}) + C

Solution steps

\int 3 1 + 2 x^{2} x^{3} d x

Apply u-substitution

: \int \frac{3 u ( u - 1 )}{8} d u

= \int \frac{3 u ( u - 1 )}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int 3 u (u - 1) d u

Expand

3 u (u - 1) : u^{\frac{4}{3}} - 3 u

= \frac{1}{8} \cdot \int u^{\frac{4}{3}} - 3 u 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}{8} (\int u^{\frac{4}{3}} d u - \int 3 u d u)

\int u^{\frac{4}{3}} d u = \frac{3}{7} u^{\frac{7}{3}}

\int 3 u d u = \frac{3}{4} u^{\frac{4}{3}}

= \frac{1}{8} (\frac{3}{7} u^{\frac{7}{3}} - \frac{3}{4} u^{\frac{4}{3}})

Substitute back

u = 1 + 2 x^{2}

= \frac{1}{8} (\frac{3}{7} (1 + 2 x^{2})^{\frac{7}{3}} - \frac{3}{4} (1 + 2 x^{2})^{\frac{4}{3}})

Add a constant to the solution

= \frac{1}{8} (\frac{3}{7} (1 + 2 x^{2})^{\frac{7}{3}} - \frac{3}{4} (1 + 2 x^{2})^{\frac{4}{3}}) + C

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What is the integral of \sqrt[3]{1+2x^2}x^3 ?
The integral of \sqrt[3]{1+2x^2}x^3 is 1/8 (3/7 (1+2x^2)^{7/3}-3/4 (1+2x^2)^{4/3})+C