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

The integral of (x^2+x-2)/(3x^3-x^2+3x-1) is -7/15 ln|3x-1|+1/5 (2ln|x^2+1|+3arctan(x))+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+x-2)/(3x^3-x^2+3x-1)

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

- \frac{7}{15} ln ∣ 3 x - 1 ∣ + \frac{1}{5} (2 ln x^{2} + 1 + 3 arctan (x)) + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} + x - 2}{3 x ^{3} - x ^{2} + 3 x - 1} : - \frac{7}{5 ( 3 x - 1 )} + \frac{4 x + 3}{5 ( x ^{2} + 1 )}

= \int - \frac{7}{5 ( 3 x - 1 )} + \frac{4 x + 3}{5 ( 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{7}{5 ( 3 x - 1 )} d x + \int \frac{4 x + 3}{5 ( x ^{2} + 1 )} d x

\int \frac{7}{5 ( 3 x - 1 )} d x = \frac{7}{15} ln ∣ 3 x - 1 ∣

\int \frac{4 x + 3}{5 ( x ^{2} + 1 )} d x = \frac{1}{5} (2 ln x^{2} + 1 + 3 arctan (x))

= - \frac{7}{15} ln ∣ 3 x - 1 ∣ + \frac{1}{5} (2 ln x^{2} + 1 + 3 arctan (x))

Add a constant to the solution

= - \frac{7}{15} ln ∣ 3 x - 1 ∣ + \frac{1}{5} (2 ln x^{2} + 1 + 3 arctan (x)) + C

\int cot^{3} (y) d y

\int \frac{3 x ^{2} - 33 x + 91}{( x - 5 ) ^{3}} d x

\int \frac{1}{x ^{2} + 7 x + 10} d x

\int \frac{cos ( ln ( x ) ) sin ( ln ( x ) )}{x} d x

\int sin^{2} (x) cos (x) e^{s i n (x)} d x

What is the integral of (x^2+x-2)/(3x^3-x^2+3x-1) ?
The integral of (x^2+x-2)/(3x^3-x^2+3x-1) is -7/15 ln|3x-1|+1/5 (2ln|x^2+1|+3arctan(x))+C