What is the integral of (x+3)/(x^2-x+6) ?

The integral of (x+3)/(x^2-x+6) is 1/2 ln|4x^2-4x+24|+7/(sqrt(23))arctan(1/(sqrt(23))(2x-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+3)/(x^2-x+6)

\int \frac{x + 3}{x ^{2} - x + 6} d x

\frac{1}{2} ln 4 x^{2} - 4 x + 24 + \frac{7}{23} arctan (\frac{1}{23} (2 x - 1)) + C

Solution steps

\int \frac{x + 3}{x ^{2} - x + 6} d x

Expand

\frac{x + 3}{x ^{2} - x + 6} : \frac{x}{x ^{2} - x + 6} + \frac{3}{x ^{2} - x + 6}

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{x}{x ^{2} - x + 6} d x + \int \frac{3}{x ^{2} - x + 6} d x

\int \frac{x}{x ^{2} - x + 6} d x = \frac{1}{2} ln 4 x^{2} - 4 x + 24 + \frac{1}{23} arctan (\frac{2 x - 1}{23})

\int \frac{3}{x ^{2} - x + 6} d x = \frac{6}{23} arctan (\frac{2 x - 1}{23})

= \frac{1}{2} ln 4 x^{2} - 4 x + 24 + \frac{1}{23} arctan (\frac{2 x - 1}{23}) + \frac{6}{23} arctan (\frac{2 x - 1}{23})

Simplify

\frac{1}{2} ln 4 x^{2} - 4 x + 24 + \frac{1}{23} arctan (\frac{2 x - 1}{23}) + \frac{6}{23} arctan (\frac{2 x - 1}{23}) : \frac{1}{2} ln 4 x^{2} - 4 x + 24 + \frac{7}{23} arctan (\frac{1}{23} (2 x - 1))

= \frac{1}{2} ln 4 x^{2} - 4 x + 24 + \frac{7}{23} arctan (\frac{1}{23} (2 x - 1))

Add a constant to the solution

= \frac{1}{2} ln 4 x^{2} - 4 x + 24 + \frac{7}{23} arctan (\frac{1}{23} (2 x - 1)) + C

\int (x + 1) (x + 1) d x

\int tan^{2} (\frac{3}{4} x) d x

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

\int \frac{cos ( x )}{2 sin ^{2} ( x ) + 3 sin ( x )} d x

\int sin^{6} (v) cos^{3} (v) d v

What is the integral of (x+3)/(x^2-x+6) ?
The integral of (x+3)/(x^2-x+6) is 1/2 ln|4x^2-4x+24|+7/(sqrt(23))arctan(1/(sqrt(23))(2x-1))+C