What is the integral of (x^2-x-21)/(2x^3-x^2+8x-4) ?

The integral of (x^2-x-21)/(2x^3-x^2+8x-4) is -5/2 ln|2x-1|+3/2 ln|x^2+4|+1/2 arctan(x/2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x^2-x-21)/(2x^3-x^2+8x-4)

\int \frac{x ^{2} - x - 21}{2 x ^{3} - x ^{2} + 8 x - 4} d x

- \frac{5}{2} ln ∣ 2 x - 1 ∣ + \frac{3}{2} ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2}) + C

Solution steps

\int \frac{x ^{2} - x - 21}{2 x ^{3} - x ^{2} + 8 x - 4} d x

Take the partial fraction of

\frac{x ^{2} - x - 21}{2 x ^{3} - x ^{2} + 8 x - 4} : - \frac{5}{2 x - 1} + \frac{3 x + 1}{x ^{2} + 4}

= \int - \frac{5}{2 x - 1} + \frac{3 x + 1}{x ^{2} + 4} 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{5}{2 x - 1} d x + \int \frac{3 x + 1}{x ^{2} + 4} d x

\int \frac{5}{2 x - 1} d x = \frac{5}{2} ln ∣ 2 x - 1 ∣

\int \frac{3 x + 1}{x ^{2} + 4} d x = \frac{3}{2} ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2})

= - \frac{5}{2} ln ∣ 2 x - 1 ∣ + \frac{3}{2} ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2})

Add a constant to the solution

= - \frac{5}{2} ln ∣ 2 x - 1 ∣ + \frac{3}{2} ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2}) + C

What is the integral of (x^2-x-21)/(2x^3-x^2+8x-4) ?
The integral of (x^2-x-21)/(2x^3-x^2+8x-4) is -5/2 ln|2x-1|+3/2 ln|x^2+4|+1/2 arctan(x/2)+C