What is the integral of 6sqrt(8x+4) ?

The integral of 6sqrt(8x+4) is 4(2x+1)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 6sqrt(8x+4)

\int 6 8 x + 4 d x

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

Solution steps

\int 6 8 x + 4 d x

Take the constant out:

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

= 6 \cdot \int 8 x + 4 d x

Apply u-substitution

= 6 \cdot \int \frac{u}{8} d u

Take the constant out:

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

= 6 \cdot \frac{1}{8} \cdot \int u d u

Apply the Power Rule

= 6 \cdot \frac{1}{8} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 8 x + 4

= 6 \cdot \frac{1}{8} \cdot \frac{2}{3} (8 x + 4)^{\frac{3}{2}}

Simplify

6 \cdot \frac{1}{8} \cdot \frac{2}{3} (8 x + 4)^{\frac{3}{2}} : 4 (2 x + 1)^{\frac{3}{2}}

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

Add a constant to the solution

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

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

\int x - a d x

\int \frac{u}{u} d u

\int (x^{\frac{3}{2}} + 3)^{4} x d x

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

What is the integral of 6sqrt(8x+4) ?
The integral of 6sqrt(8x+4) is 4(2x+1)^{3/2}+C