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

The integral of (x^2-2x-1)/(2x^3+3x^2-2x) is 1/2 ln|x|-7/10 ln|2x-1|+7/10 ln|x+2|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{2} ln ∣ x ∣ - \frac{7}{10} ln ∣ 2 x - 1 ∣ + \frac{7}{10} ln ∣ x + 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} - 2 x - 1}{2 x ^{3} + 3 x ^{2} - 2 x} : \frac{1}{2 x} - \frac{7}{5 ( 2 x - 1 )} + \frac{7}{10 ( x + 2 )}

= \int \frac{1}{2 x} - \frac{7}{5 ( 2 x - 1 )} + \frac{7}{10 ( x + 2 )} 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{7}{5 ( 2 x - 1 )} d x + \int \frac{7}{10 ( x + 2 )} d x

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

\int \frac{7}{5 ( 2 x - 1 )} d x = \frac{7}{10} ln ∣ 2 x - 1 ∣

\int \frac{7}{10 ( x + 2 )} d x = \frac{7}{10} ln ∣ x + 2 ∣

= \frac{1}{2} ln ∣ x ∣ - \frac{7}{10} ln ∣ 2 x - 1 ∣ + \frac{7}{10} ln ∣ x + 2 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ x ∣ - \frac{7}{10} ln ∣ 2 x - 1 ∣ + \frac{7}{10} ln ∣ x + 2 ∣ + C

\int \frac{x ^{2}}{e ^{5 x}} d x

\frac{8 e ^{(- 3 s)}}{s ^{3}}

\frac{1}{1 - y}

\frac{d}{d x} (3 x^{2} cos (x) cot (x))

x = 6 y^{2}, x = 16 + 2 y^{2}

What is the integral of (x^2-2x-1)/(2x^3+3x^2-2x) ?
The integral of (x^2-2x-1)/(2x^3+3x^2-2x) is 1/2 ln|x|-7/10 ln|2x-1|+7/10 ln|x+2|+C