What is the integral of (2x+1)/(x^2-6x+13) ?

The integral of (2x+1)/(x^2-6x+13) is ln|x^2-6x+13|+3arctan(1/2 (x-3))+1/2 arctan((x-3)/2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (2x+1)/(x^2-6x+13)

\int \frac{2 x + 1}{x ^{2} - 6 x + 13} d x

ln x^{2} - 6 x + 13 + 3 arctan (\frac{1}{2} (x - 3)) + \frac{1}{2} arctan (\frac{x - 3}{2}) + C

Solution steps

\int \frac{2 x + 1}{x ^{2} - 6 x + 13} d x

Simplify

x^{2} : xx

= \int \frac{2 x + 1}{xx - 6 x + 13} d x

Expand

\frac{2 x + 1}{xx - 6 x + 13} : \frac{2 x}{xx - 6 x + 13} + \frac{1}{xx - 6 x + 13}

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{2 x}{xx - 6 x + 13} d x + \int \frac{1}{xx - 6 x + 13} d x

\int \frac{2 x}{xx - 6 x + 13} d x = ln x^{2} - 6 x + 13 + 3 arctan (\frac{1}{2} (x - 3))

\int \frac{1}{xx - 6 x + 13} d x = \frac{1}{2} arctan (\frac{x - 3}{2})

= ln x^{2} - 6 x + 13 + 3 arctan (\frac{1}{2} (x - 3)) + \frac{1}{2} arctan (\frac{x - 3}{2})

Add a constant to the solution

= ln x^{2} - 6 x + 13 + 3 arctan (\frac{1}{2} (x - 3)) + \frac{1}{2} arctan (\frac{x - 3}{2}) + C

What is the integral of (2x+1)/(x^2-6x+13) ?
The integral of (2x+1)/(x^2-6x+13) is ln|x^2-6x+13|+3arctan(1/2 (x-3))+1/2 arctan((x-3)/2)+C