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

The integral of (-x^2-x+1)/(4x^3-2x^2-2x) is -1/2 ln|x|+5/12 ln|2x+1|-1/6 ln|x-1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (-x^2-x+1)/(4x^3-2x^2-2x)

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

- \frac{1}{2} ln ∣ x ∣ + \frac{5}{12} ln ∣ 2 x + 1 ∣ - \frac{1}{6} ln ∣ x - 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{- x ^{2} - x + 1}{4 x ^{3} - 2 x ^{2} - 2 x} : - \frac{1}{2 x} + \frac{5}{6 ( 2 x + 1 )} - \frac{1}{6 ( x - 1 )}

= \int - \frac{1}{2 x} + \frac{5}{6 ( 2 x + 1 )} - \frac{1}{6 ( 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}{2 x} d x + \int \frac{5}{6 ( 2 x + 1 )} d x - \int \frac{1}{6 ( x - 1 )} d x

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

\int \frac{5}{6 ( 2 x + 1 )} d x = \frac{5}{12} ln ∣ 2 x + 1 ∣

\int \frac{1}{6 ( x - 1 )} d x = \frac{1}{6} ln ∣ x - 1 ∣

= - \frac{1}{2} ln ∣ x ∣ + \frac{5}{12} ln ∣ 2 x + 1 ∣ - \frac{1}{6} ln ∣ x - 1 ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ x ∣ + \frac{5}{12} ln ∣ 2 x + 1 ∣ - \frac{1}{6} ln ∣ x - 1 ∣ + C

\int x^{2} \cdot e^{(x^{3})} d x

\int \frac{1}{( - 4 x ^{2} - 20 x - 21 )} d x

\int \frac{sin ( 3 x )}{3 x} d x

\int \frac{29 - 51 x ^{3}}{x} d x

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

What is the integral of (-x^2-x+1)/(4x^3-2x^2-2x) ?
The integral of (-x^2-x+1)/(4x^3-2x^2-2x) is -1/2 ln|x|+5/12 ln|2x+1|-1/6 ln|x-1|+C