What is the integral of 18x^2\sqrt[4]{6x^3+5} ?

The integral of 18x^2\sqrt[4]{6x^3+5} is 4/5 (6x^3+5)^{5/4}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 18x^2\sqrt[4]{6x^3+5}

\int 18 x^{2} 4 6 x^{3} + 5 d x

\frac{4}{5} (6 x^{3} + 5)^{\frac{5}{4}} + C

Solution steps

\int 18 x^{2} 4 6 x^{3} + 5 d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 18 \cdot \int x^{2} 4 6 x^{3} + 5 d x

Apply u-substitution

= 18 \cdot \int \frac{4 u}{18} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 18 \cdot \frac{1}{18} \cdot \int 4 u d u

Apply the Power Rule

= 18 \cdot \frac{1}{18} \cdot \frac{4}{5} u^{\frac{5}{4}}

Substitute back

u = 6 x^{3} + 5

= 18 \cdot \frac{1}{18} \cdot \frac{4}{5} (6 x^{3} + 5)^{\frac{5}{4}}

Simplify

18 \cdot \frac{1}{18} \cdot \frac{4}{5} (6 x^{3} + 5)^{\frac{5}{4}} : \frac{4}{5} (6 x^{3} + 5)^{\frac{5}{4}}

= \frac{4}{5} (6 x^{3} + 5)^{\frac{5}{4}}

Add a constant to the solution

= \frac{4}{5} (6 x^{3} + 5)^{\frac{5}{4}} + C

\int \frac{1}{x ^{2} + \frac{1}{4}} d x

\int \frac{1}{9} cos (x) d x

\int 10 x (5 x^{2} + 9)^{7} d x

\int 4 x^{3} + 3 x + \frac{1}{x ^{3}} d x

\int sin^{2} (a)

What is the integral of 18x^2\sqrt[4]{6x^3+5} ?
The integral of 18x^2\sqrt[4]{6x^3+5} is 4/5 (6x^3+5)^{5/4}+C