What is the integral of 2x\sqrt[3]{x^2-5} ?

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

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integral of 2x\sqrt[3]{x^2-5}

\frac{3}{4} (x^{2} - 5)^{\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 = x^{2} - 5

= 2 \cdot \frac{1}{2} \cdot \frac{3}{4} (x^{2} - 5)^{\frac{4}{3}}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{3}{4} (x^{2} - 5)^{\frac{4}{3}} : \frac{3}{4} (x^{2} - 5)^{\frac{4}{3}}

= \frac{3}{4} (x^{2} - 5)^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{4} (x^{2} - 5)^{\frac{4}{3}} + C

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