What is the integral of (x-2)/(x^2-2x+5) ?

The integral of (x-2)/(x^2-2x+5) is 1/2 ln|x^2-2x+5|-1/2 arctan(1/2 (x-1))+C

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integral of (x-2)/(x^2-2x+5)

\int \frac{x - 2}{x ^{2} - 2 x + 5} d x

\frac{1}{2} ln x^{2} - 2 x + 5 - \frac{1}{2} arctan (\frac{1}{2} (x - 1)) + C

Solution steps

\int \frac{x - 2}{x ^{2} - 2 x + 5} d x

Expand

\frac{x - 2}{x ^{2} - 2 x + 5} : \frac{x}{x ^{2} - 2 x + 5} - \frac{2}{x ^{2} - 2 x + 5}

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 + 5} d x - \int \frac{2}{x ^{2} - 2 x + 5} d x

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

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

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

Add similar elements:

\frac{1}{2} arctan (\frac{1}{2} (x - 1)) - arctan (\frac{1}{2} (x - 1)) = - \frac{1}{2} arctan (\frac{1}{2} (x - 1))

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

Add a constant to the solution

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

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What is the integral of (x-2)/(x^2-2x+5) ?
The integral of (x-2)/(x^2-2x+5) is 1/2 ln|x^2-2x+5|-1/2 arctan(1/2 (x-1))+C