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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

integral

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

Add a constant to the solution

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

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

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

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

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

- sin^{2} (x) + tan^{2} (x)

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