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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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{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

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