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

The integral of (x^2)/(\sqrt[4]{x^3+4)} is 4/9 (x^3+4)^{3/4}+C

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

\frac{4}{9} (x^{3} + 4)^{\frac{3}{4}} + 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{4}{3} u^{\frac{3}{4}}

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

Substitute back

u = x^{3} + 4

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

Simplify

\frac{1}{3} \cdot \frac{4}{3} (x^{3} + 4)^{\frac{3}{4}} : \frac{4}{9} (x^{3} + 4)^{\frac{3}{4}}

= \frac{4}{9} (x^{3} + 4)^{\frac{3}{4}}

Add a constant to the solution

= \frac{4}{9} (x^{3} + 4)^{\frac{3}{4}} + C

x^{3} - 8 x^{2} + 19 x - 12

\int_{5}^{10} (3 x + 3) d x

\frac{\partial}{\partial y} (x^{3} - y^{2} + 2 x y)

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

\frac{d y}{d x} = 8 y e^{x + 2}

What is the integral of (x^2)/(\sqrt[4]{x^3+4)} ?
The integral of (x^2)/(\sqrt[4]{x^3+4)} is 4/9 (x^3+4)^{3/4}+C