What is the integral of (2x^2)/(2x-3) ?

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

\int \frac{2 x ^{2}}{2 x - 3} d x

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

Solution steps

\int \frac{2 x ^{2}}{2 x - 3} 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 ^{2}}{2 x - 3} d x

Long division

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

= 2 \cdot \int \frac{x}{2} + \frac{3}{4} + \frac{9}{4 ( 2 x - 3 )} 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

= 2 (\int \frac{x}{2} d x + \int \frac{3}{4} d x + \int \frac{9}{4 ( 2 x - 3 )} d x)

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

\int \frac{3}{4} d x = \frac{3}{4} x

\int \frac{9}{4 ( 2 x - 3 )} d x = \frac{9}{8} ln ∣ 2 x - 3 ∣

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

Add a constant to the solution

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

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

\frac{\partial}{\partial x} ((x + y)^{2})

\int 2 (2 x + 8)^{4} d x

\int_{0}^{π} 4 x cos (x^{2}) d x

\frac{d x}{d t} = x - 1

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