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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

\int \frac{4 x + 3}{x ^{2} - 6 x + 13} d x

\int 5 x^{4} (2 x^{5} - 3)^{8} d x

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

\int 36 z^{2} (216 z^{3} + 1)^{12} d z

\int (- 9 x^{2} - 4 x + 1) d x

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