What is the integral of (x-4)/(x^2-2x+4) ?

The integral of (x-4)/(x^2-2x+4) is 1/2 ln|x^2-2x+4|-sqrt(3)arctan(1/(sqrt(3))(x-1))+C

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integral of (x-4)/(x^2-2x+4)

\int \frac{x - 4}{x ^{2} - 2 x + 4} d x

\frac{1}{2} ln x^{2} - 2 x + 4 - 3 arctan (\frac{1}{3} (x - 1)) + C

Solution steps

\int \frac{x - 4}{x ^{2} - 2 x + 4} d x

Expand

\frac{x - 4}{x ^{2} - 2 x + 4} : \frac{x}{x ^{2} - 2 x + 4} - \frac{4}{x ^{2} - 2 x + 4}

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} - 2 x + 4} d x - \int \frac{4}{x ^{2} - 2 x + 4} d x

\int \frac{x}{x ^{2} - 2 x + 4} d x = \frac{1}{2} ln x^{2} - 2 x + 4 + \frac{1}{3} arctan (\frac{1}{3} (x - 1))

\int \frac{4}{x ^{2} - 2 x + 4} d x = \frac{4}{3} arctan (\frac{1}{3} (x - 1))

= \frac{1}{2} ln x^{2} - 2 x + 4 + \frac{1}{3} arctan (\frac{1}{3} (x - 1)) - \frac{4}{3} arctan (\frac{1}{3} (x - 1))

Add similar elements:

\frac{1}{3} arctan (\frac{1}{3} (x - 1)) - \frac{4}{3} arctan (\frac{1}{3} (x - 1)) = - 3 arctan (\frac{1}{3} (x - 1))

= \frac{1}{2} ln x^{2} - 2 x + 4 - 3 arctan (\frac{1}{3} (x - 1))

Add a constant to the solution

= \frac{1}{2} ln x^{2} - 2 x + 4 - 3 arctan (\frac{1}{3} (x - 1)) + C

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