What is the integral of (4x)/(sqrt(2x+3)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (4x)/(sqrt(2x+3))

\int \frac{4 x}{2 x + 3} d x

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

Solution steps

\int \frac{4 x}{2 x + 3} d x

Take the constant out:

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

= 4 \cdot \int \frac{x}{2 x + 3} d x

Apply u-substitution

: \int \frac{u ^{2} - 3}{2} d u

= 4 \cdot \int \frac{u ^{2} - 3}{2} d u

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int u^{2} - 3 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

= 4 \cdot \frac{1}{2} (\int u^{2} d u - \int 3 d u)

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

\int 3 d u = 3 u

= 4 \cdot \frac{1}{2} (\frac{u ^{3}}{3} - 3 u)

Substitute back

u = 2 x + 3

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

Simplify

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

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

Add a constant to the solution

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

f (x) = 3 - 6 x

x \to 1 lim (- x + 2)

\frac{\partial}{\partial x} (x + 3 \cdot ln (x) - y^{2})

\frac{d}{d x} (a x^{2} + x)

\int_{5}^{6} \frac{1}{x ^{2} - 1} d x

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