What is the integral of 1/(2x^2-x) ?

The integral of 1/(2x^2-x) is -ln|x|+ln|2x-1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(2x^2-x)

\int \frac{1}{2 x ^{2} - x} d x

- ln ∣ x ∣ + ln ∣ 2 x - 1 ∣ + C

Solution steps

\int \frac{1}{2 x ^{2} - x} d x

Take the partial fraction of

\frac{1}{2 x ^{2} - x} : - \frac{1}{x} + \frac{2}{2 x - 1}

= \int - \frac{1}{x} + \frac{2}{2 x - 1} 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

= - \int \frac{1}{x} d x + \int \frac{2}{2 x - 1} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{2}{2 x - 1} d x = ln ∣ 2 x - 1 ∣

= - ln ∣ x ∣ + ln ∣ 2 x - 1 ∣

Add a constant to the solution

= - ln ∣ x ∣ + ln ∣ 2 x - 1 ∣ + C

f (x) = 5 x - 7

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

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

y = - 9 x^{2}, at x = 8

y^{^{''}} = x^{2} + 1

What is the integral of 1/(2x^2-x) ?
The integral of 1/(2x^2-x) is -ln|x|+ln|2x-1|+C