What is the integral of (15x\sqrt[4]{3x^2+4})/4 ?

The integral of (15x\sqrt[4]{3x^2+4})/4 is 1/2 (3x^2+4)^{5/4}+C

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integral of (15x\sqrt[4]{3x^2+4})/4

\frac{1}{2} (3 x^{2} + 4)^{\frac{5}{4}} + 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

: \frac{4}{5} u^{\frac{5}{4}}

= \frac{15}{4} \cdot \frac{1}{6} \cdot \frac{4}{5} u^{\frac{5}{4}}

Substitute back

u = 3 x^{2} + 4

= \frac{15}{4} \cdot \frac{1}{6} \cdot \frac{4}{5} (3 x^{2} + 4)^{\frac{5}{4}}

Simplify

\frac{15}{4} \cdot \frac{1}{6} \cdot \frac{4}{5} (3 x^{2} + 4)^{\frac{5}{4}} : \frac{1}{2} (3 x^{2} + 4)^{\frac{5}{4}}

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

Add a constant to the solution

= \frac{1}{2} (3 x^{2} + 4)^{\frac{5}{4}} + C

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What is the integral of (15x\sqrt[4]{3x^2+4})/4 ?
The integral of (15x\sqrt[4]{3x^2+4})/4 is 1/2 (3x^2+4)^{5/4}+C