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

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

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

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

Solution steps

Take the constant out:

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

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

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

Substitute back

u = 1 + x^{2}

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

Simplify

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

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

Add a constant to the solution

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

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