What is the integral of x\sqrt[4]{4x^2+5} ?

The integral of x\sqrt[4]{4x^2+5} is 1/10 (4x^2+5)^{5/4}+C

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integral of x\sqrt[4]{4x^2+5}

\frac{1}{10} (4 x^{2} + 5)^{\frac{5}{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}{5} u^{\frac{5}{4}}

= \frac{1}{8} \cdot \frac{4}{5} u^{\frac{5}{4}}

Substitute back

u = 4 x^{2} + 5

= \frac{1}{8} \cdot \frac{4}{5} (4 x^{2} + 5)^{\frac{5}{4}}

Simplify

\frac{1}{8} \cdot \frac{4}{5} (4 x^{2} + 5)^{\frac{5}{4}} : \frac{1}{10} (4 x^{2} + 5)^{\frac{5}{4}}

= \frac{1}{10} (4 x^{2} + 5)^{\frac{5}{4}}

Add a constant to the solution

= \frac{1}{10} (4 x^{2} + 5)^{\frac{5}{4}} + C

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What is the integral of x\sqrt[4]{4x^2+5} ?
The integral of x\sqrt[4]{4x^2+5} is 1/10 (4x^2+5)^{5/4}+C