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

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

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

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

Solution steps

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

Take the partial fraction of

\frac{x}{x ^{3} + x ^{2} - x - 1} : - \frac{1}{4 ( x + 1 )} + \frac{1}{2 ( x + 1 ) ^{2}} + \frac{1}{4 ( x - 1 )}

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

\int \frac{1}{4 ( x + 1 )} d x = \frac{1}{4} ln ∣ x + 1 ∣

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

\int \frac{1}{4 ( x - 1 )} d x = \frac{1}{4} ln ∣ x - 1 ∣

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

Add a constant to the solution

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

\int - \frac{6}{7 x 49 x ^{2} - 25} d x

\int \frac{( y ^{3} - y ^{2} + 1 )}{( y ^{2} + 2 y + 2 ) ^{2}} d y

\int cos (\frac{π}{2} x^{2}) d x

\int sec (2 z) tan (2 z) d z

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