What is the integral of 6x^2sqrt((2x^3)) ?

The integral of 6x^2sqrt((2x^3)) is (4sqrt(2))/3 x^{9/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 6x^2sqrt((2x^3))

\int 6 x^{2} (2 x^{3}) d x

\frac{4 2}{3} x^{\frac{9}{2}} + C

Solution steps

\int 6 x^{2} 2 x^{3} d x

Take the constant out:

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

= 6 \cdot \int x^{2} 2 x^{3} d x

Apply u-substitution

: \int \frac{u ^{\frac{5}{4}}}{2} d u

= 6 \cdot \int \frac{u ^{\frac{5}{4}}}{2} d u

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \int u^{\frac{5}{4}} d u

Apply the Power Rule

: \frac{4}{9} u^{\frac{9}{4}}

= 6 \cdot \frac{1}{2} \cdot \frac{4}{9} u^{\frac{9}{4}}

Substitute back

u = x^{2}

= 6 \cdot \frac{1}{2} \cdot \frac{4}{9} (x^{2})^{\frac{9}{4}}

Simplify

6 \cdot \frac{1}{2} \cdot \frac{4}{9} (x^{2})^{\frac{9}{4}} : \frac{4 2}{3} x^{\frac{9}{2}}

= \frac{4 2}{3} x^{\frac{9}{2}}

Add a constant to the solution

= \frac{4 2}{3} x^{\frac{9}{2}} + C

\int \frac{6 x ^{2} - 3 x + 5}{( 2 x - 1 ) ( x ^{2} + 1 )} d x

\int \frac{2 x ^{2}}{x + 6} d x

\int sin^{2} (x) + 3 d x

\int \frac{1}{x ^{4} 4 x ^{2} - 9} d x

\int - \frac{2}{81 + 25 t ^{2}} d t

What is the integral of 6x^2sqrt((2x^3)) ?
The integral of 6x^2sqrt((2x^3)) is (4sqrt(2))/3 x^{9/2}+C