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

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integral of ((x+3))/((x^2-x+6))

integral

\frac{( x + 3 )}{( x ^{2} - x + 6 )}

\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

h (x) = \frac{4 x ^{3} + 2 x + 5}{x}

\frac{t ^{2}}{2} + \frac{t ^{3}}{3} + \frac{t ^{4}}{4}

(\frac{sin ( x )}{x})^{2}

sinh (5 x + 2)

\frac{- 4}{( x ^{2} + 4 ) ^{2}}

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