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|+1/10 ln|2x-1|-1/10 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{1}{10} ln ∣ 2 x - 1 ∣ - \frac{1}{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{1}{5 ( 2 x - 1 )} - \frac{1}{10 ( x + 2 )}

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

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

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

\int \frac{1}{10 ( x + 2 )} d x = \frac{1}{10} ln ∣ x + 2 ∣

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

Add a constant to the solution

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

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

\int x^{3} cos (k x) d x

\int \frac{( 3 x - 2 ) ^{4}}{4 x} d x

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

\int (x^{1.9} + 7 x^{2.5}) d x

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|+1/10 ln|2x-1|-1/10 ln|x+2|+C