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

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

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

\frac{3}{8} (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

Apply the Power Rule

: \frac{3}{4} u^{\frac{4}{3}}

= \frac{1}{2} \cdot \frac{3}{4} u^{\frac{4}{3}}

Substitute back

u = 1 + x^{2}

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

Simplify

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

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

Add a constant to the solution

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

n = 0 \sum \infty 8 (\frac{1}{2})^{n - 1}

x^{2} y

\int 15 sin^{2} (θ) d θ

x \to 3 lim (\frac{3}{( x - 3 ) ^{2}})

\frac{d}{d x} (ln (\frac{x ^{2}}{5} + \frac{9}{x}))

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