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

The integral of (2x)/(sqrt(3x+1)) is 4/9 (1/3 (3x+1)^{3/2}-sqrt(3x+1))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (2x)/(sqrt(3x+1))

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

\frac{4}{9} (\frac{1}{3} (3 x + 1)^{\frac{3}{2}} - 3 x + 1) + C

Solution steps

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

Take the constant out:

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

= 2 \cdot \int \frac{x}{3 x + 1} d x

Apply u-substitution

= 2 \cdot \int \frac{2 ( u ^{2} - 1 )}{9} d u

Take the constant out:

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

= 2 \cdot \frac{2}{9} \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

= 2 \cdot \frac{2}{9} (\int u^{2} d u - \int 1 d u)

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

\int 1 d u = u

= 2 \cdot \frac{2}{9} (\frac{u ^{3}}{3} - u)

Substitute back

u = 3 x + 1

= 2 \cdot \frac{2}{9} (\frac{( 3 x + 1 ) ^{3}}{3} - 3 x + 1)

Simplify

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

= \frac{4}{9} (\frac{1}{3} (3 x + 1)^{\frac{3}{2}} - 3 x + 1)

Add a constant to the solution

= \frac{4}{9} (\frac{1}{3} (3 x + 1)^{\frac{3}{2}} - 3 x + 1) + C

What is the integral of (2x)/(sqrt(3x+1)) ?
The integral of (2x)/(sqrt(3x+1)) is 4/9 (1/3 (3x+1)^{3/2}-sqrt(3x+1))+C