What is the integral of 6x^2sqrt(2x^3+1) ?

The integral of 6x^2sqrt(2x^3+1) is 2/3 (2x^3+1)^{3/2}+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+1)

\int 6 x^{2} 2 x^{3} + 1 d x

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

Solution steps

\int 6 x^{2} 2 x^{3} + 1 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} + 1 d x

Apply u-substitution

: \int \frac{u}{6} d u

= 6 \cdot \int \frac{u}{6} d u

Take the constant out:

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

= 6 \cdot \frac{1}{6} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= 6 \cdot \frac{1}{6} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 2 x^{3} + 1

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

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d t} = 4 y, y (5) = 9

\int \frac{x + 7}{x ^{2} + 16} d x

x \to - 4 lim (\frac{1}{x + 4})

f (x) = (3 x + 2)^{2}

\frac{d}{d x} (2 sin (x) + cos (2 x))

What is the integral of 6x^2sqrt(2x^3+1) ?
The integral of 6x^2sqrt(2x^3+1) is 2/3 (2x^3+1)^{3/2}+C