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

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

integral

\frac{( 2 x ^{2} - 1 )}{( x - x ^{3} )}

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Add a constant to the solution

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

\frac{sin ^{2} ( x )}{x}

\frac{3}{x ^{2} - x - 1}

2 t (9 - 4 t^{2})^{\frac{1}{2}}

- 4 x

x^{\frac{1}{6}}

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