What is the integral of (8x^2)/(4x+6) ?

The integral of (8x^2)/(4x+6) is x^2-3x+9/2 ln|4x+6|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (8x^2)/(4x+6)

\int \frac{8 x ^{2}}{4 x + 6} d x

x^{2} - 3 x + \frac{9}{2} ln ∣ 4 x + 6 ∣ + C

Solution steps

\int \frac{8 x ^{2}}{4 x + 6} d x

Take the constant out:

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

= 8 \cdot \int \frac{x ^{2}}{4 x + 6} d x

Long division

\frac{x ^{2}}{4 x + 6} : \frac{x}{4} - \frac{3}{8} + \frac{9}{4 ( 4 x + 6 )}

= 8 \cdot \int \frac{x}{4} - \frac{3}{8} + \frac{9}{4 ( 4 x + 6 )} d x

Apply the Sum Rule:

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

= 8 (\int \frac{x}{4} d x - \int \frac{3}{8} d x + \int \frac{9}{4 ( 4 x + 6 )} d x)

\int \frac{x}{4} d x = \frac{x ^{2}}{8}

\int \frac{3}{8} d x = \frac{3}{8} x

\int \frac{9}{4 ( 4 x + 6 )} d x = \frac{9}{16} ln ∣ 4 x + 6 ∣

= 8 (\frac{x ^{2}}{8} - \frac{3}{8} x + \frac{9}{16} ln ∣ 4 x + 6 ∣)

Simplify

8 (\frac{x ^{2}}{8} - \frac{3}{8} x + \frac{9}{16} ln ∣ 4 x + 6 ∣) : x^{2} - 3 x + \frac{9}{2} ln ∣ 4 x + 6 ∣

= x^{2} - 3 x + \frac{9}{2} ln ∣ 4 x + 6 ∣

Add a constant to the solution

= x^{2} - 3 x + \frac{9}{2} ln ∣ 4 x + 6 ∣ + C

\int 9 t^{4} + 4 t^{2} d t

\frac{\partial}{\partial x} (x + y - 320)

x \to 4 lim (x - 2)

\frac{d}{d x} (e^{x^{2}} y) = x

\int_{0}^{7} x + 1 d x

What is the integral of (8x^2)/(4x+6) ?
The integral of (8x^2)/(4x+6) is x^2-3x+9/2 ln|4x+6|+C