What is the integral of (3+3x-2x^2)/(5-7x+3x^2-x^3) ?

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

\int \frac{3 + 3 x - 2 x ^{2}}{5 - 7 x + 3 x ^{2} - x ^{3}} d x

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

Solution steps

\int \frac{3 + 3 x - 2 x ^{2}}{5 - 7 x + 3 x ^{2} - x ^{3}} d x

Take the partial fraction of

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

= \int - \frac{1}{x - 1} + \frac{3 x - 2}{x ^{2} - 2 x + 5} 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 - 1} d x + \int \frac{3 x - 2}{x ^{2} - 2 x + 5} d x

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

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

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

Add a constant to the solution

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

\int \frac{8 x}{4 x ^{2} - 5} d x

\int (cos (x) - 1)^{2} d x

\int (sin (x))^{n} \cdot cos (x) d x

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

\int \frac{3 x - 2}{4 - 5 x} d x

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