What is the integral of (4x^2-2x+1)/(4x^3-x) ?

The integral of (4x^2-2x+1)/(4x^3-x) is -ln|x|+3/2 ln|2x+1|+1/2 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 (4x^2-2x+1)/(4x^3-x)

\int \frac{4 x ^{2} - 2 x + 1}{4 x ^{3} - x} d x

- ln ∣ x ∣ + \frac{3}{2} ln ∣ 2 x + 1 ∣ + \frac{1}{2} ln ∣ 2 x - 1 ∣ + C

Solution steps

\int \frac{4 x ^{2} - 2 x + 1}{4 x ^{3} - x} d x

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

= - ln ∣ x ∣ + \frac{3}{2} ln ∣ 2 x + 1 ∣ + \frac{1}{2} ln ∣ 2 x - 1 ∣

Add a constant to the solution

= - ln ∣ x ∣ + \frac{3}{2} ln ∣ 2 x + 1 ∣ + \frac{1}{2} ln ∣ 2 x - 1 ∣ + C

\frac{d}{d x} (\frac{2 x ^{2} + 4 x + 4}{x})

\frac{d}{d x} (\frac{1 - 4 x - x ^{2}}{x ^{2} - 5})

\int sin^{3} (2 x) cos^{7} (x) d x

\int \frac{1}{x ^{2} - 6 x - 7} d x

x \to 0 lim (\frac{sin ( 5 x )}{x} + 4)

What is the integral of (4x^2-2x+1)/(4x^3-x) ?
The integral of (4x^2-2x+1)/(4x^3-x) is -ln|x|+3/2 ln|2x+1|+1/2 ln|2x-1|+C