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

The integral of (x^4-x+1)/(x^4(2x-1)) is -4ln|x|+2/x+1/(2x^2)+1/(3x^3)+9/2 ln|2x-1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x^4-x+1)/(x^4(2x-1))

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

- 4 ln ∣ x ∣ + \frac{2}{x} + \frac{1}{2 x ^{2}} + \frac{1}{3 x ^{3}} + \frac{9}{2} ln ∣ 2 x - 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{4} - x + 1}{x ^{4} ( 2 x - 1 )} : - \frac{4}{x} - \frac{2}{x ^{2}} - \frac{1}{x ^{3}} - \frac{1}{x ^{4}} + \frac{9}{2 x - 1}

= \int - \frac{4}{x} - \frac{2}{x ^{2}} - \frac{1}{x ^{3}} - \frac{1}{x ^{4}} + \frac{9}{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{4}{x} d x - \int \frac{2}{x ^{2}} d x - \int \frac{1}{x ^{3}} d x - \int \frac{1}{x ^{4}} d x + \int \frac{9}{2 x - 1} d x

\int \frac{4}{x} d x = 4 ln ∣ x ∣

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

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

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

\int \frac{9}{2 x - 1} d x = \frac{9}{2} ln ∣ 2 x - 1 ∣

= - 4 ln ∣ x ∣ - (- \frac{2}{x}) - (- \frac{1}{2 x ^{2}}) - (- \frac{1}{3 x ^{3}}) + \frac{9}{2} ln ∣ 2 x - 1 ∣

Simplify

= - 4 ln ∣ x ∣ + \frac{2}{x} + \frac{1}{2 x ^{2}} + \frac{1}{3 x ^{3}} + \frac{9}{2} ln ∣ 2 x - 1 ∣

Add a constant to the solution

= - 4 ln ∣ x ∣ + \frac{2}{x} + \frac{1}{2 x ^{2}} + \frac{1}{3 x ^{3}} + \frac{9}{2} ln ∣ 2 x - 1 ∣ + C

What is the integral of (x^4-x+1)/(x^4(2x-1)) ?
The integral of (x^4-x+1)/(x^4(2x-1)) is -4ln|x|+2/x+1/(2x^2)+1/(3x^3)+9/2 ln|2x-1|+C