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

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

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

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

2 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)) - \frac{3}{5} (ln \frac{2}{5} x + \frac{6}{5} - ln \frac{2}{5} x - \frac{4}{5}) + 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 = 2 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x))

\int \frac{3}{x ^{2} + x - 6} d x = - \frac{3}{5} (ln \frac{2}{5} x + \frac{6}{5} - ln \frac{2}{5} x - \frac{4}{5})

= 2 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)) - \frac{3}{5} (ln \frac{2}{5} x + \frac{6}{5} - ln \frac{2}{5} x - \frac{4}{5})

Add a constant to the solution

= 2 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)) - \frac{3}{5} (ln \frac{2}{5} x + \frac{6}{5} - ln \frac{2}{5} x - \frac{4}{5}) + C

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