What is the integral of (12x)/(sqrt(1+4x)) ?

The integral of (12x)/(sqrt(1+4x)) is (2x-1)sqrt(4x+1)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (12x)/(sqrt(1+4x))

\int \frac{12 x}{1 + 4 x} d x

(2 x - 1) 4 x + 1 + C

Solution steps

\int \frac{12 x}{1 + 4 x} d x

Take the constant out:

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

= 12 \cdot \int \frac{x}{1 + 4 x} d x

Apply u-substitution

: \int \frac{u ^{2} - 1}{8} d u

= 12 \cdot \int \frac{u ^{2} - 1}{8} d u

Take the constant out:

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

= 12 \cdot \frac{1}{8} \cdot \int u^{2} - 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 12 \cdot \frac{1}{8} (\int u^{2} d u - \int 1 d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int 1 d u = u

= 12 \cdot \frac{1}{8} (\frac{u ^{3}}{3} - u)

Substitute back

u = 1 + 4 x

= 12 \cdot \frac{1}{8} (\frac{( 1 + 4 x ) ^{3}}{3} - 1 + 4 x)

Simplify

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

= (2 x - 1) 4 x + 1

Add a constant to the solution

= (2 x - 1) 4 x + 1 + C

\frac{d}{d x} (\frac{x ^{3}}{3} + x^{2} + 4 x - 10)

\int sin (2 π)

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

\frac{\partial}{\partial x} (x y)

\frac{d}{d x} (ln (0))

What is the integral of (12x)/(sqrt(1+4x)) ?
The integral of (12x)/(sqrt(1+4x)) is (2x-1)sqrt(4x+1)+C