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

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

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

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

\frac{3}{2} ln x^{2} - 2 x + 6 + C

Solution steps

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

Expand

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

\int \frac{3 x}{x ^{2} - 2 x + 6} d x = 3 (\frac{1}{2} ln x^{2} - 2 x + 6 + \frac{1}{5} arctan (\frac{1}{5} (x - 1)))

\int \frac{3}{x ^{2} - 2 x + 6} d x = \frac{3}{5} arctan (\frac{1}{5} (x - 1))

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

Simplify

3 (\frac{1}{2} ln x^{2} - 2 x + 6 + \frac{1}{5} arctan (\frac{1}{5} (x - 1))) - \frac{3}{5} arctan (\frac{1}{5} (x - 1)) : \frac{3}{2} ln x^{2} - 2 x + 6

= \frac{3}{2} ln x^{2} - 2 x + 6

Add a constant to the solution

= \frac{3}{2} ln x^{2} - 2 x + 6 + C

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