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

The integral of (2x^2+4x-1)/(x^3-x^2+4x-4) is ln|x-1|+1/2 ln|x^2+4|+5/2 arctan(x/2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x^2+4x-1)/(x^3-x^2+4x-4)

\int \frac{2 x ^{2} + 4 x - 1}{x ^{3} - x ^{2} + 4 x - 4} d x

ln ∣ x - 1 ∣ + \frac{1}{2} ln x^{2} + 4 + \frac{5}{2} arctan (\frac{x}{2}) + C

Solution steps

\int \frac{2 x ^{2} + 4 x - 1}{x ^{3} - x ^{2} + 4 x - 4} d x

Take the partial fraction of

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

= \int \frac{1}{x - 1} + \frac{x + 5}{x ^{2} + 4} 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{x + 5}{x ^{2} + 4} d x

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

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

= ln ∣ x - 1 ∣ + \frac{1}{2} ln x^{2} + 4 + \frac{5}{2} arctan (\frac{x}{2})

Add a constant to the solution

= ln ∣ x - 1 ∣ + \frac{1}{2} ln x^{2} + 4 + \frac{5}{2} arctan (\frac{x}{2}) + C

\int \frac{x ^{2} - 9 x + 25}{x ^{2} - 9 x + 20} d x

\int e^{y - 1} d y

\int (8 x^{5} - 5 x^{4} - 3 x^{3} + 6 x^{2} - 10 x - 5) d x

\int \frac{7 x}{( 6 - x ^{2} ) ^{3}} d x

\int 3 sin (x) e^{x} d x

What is the integral of (2x^2+4x-1)/(x^3-x^2+4x-4) ?
The integral of (2x^2+4x-1)/(x^3-x^2+4x-4) is ln|x-1|+1/2 ln|x^2+4|+5/2 arctan(x/2)+C