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

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

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

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

Expand

Apply the Sum Rule:

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

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

= \frac{1}{2} (\frac{3}{7} u^{\frac{7}{3}} - \frac{3}{4} u^{\frac{4}{3}})

Substitute back

u = 1 + x^{2}

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

Add a constant to the solution

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

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