What is the integral of ((5x-6))/((x^2-2x+2)) ?

The integral of ((5x-6))/((x^2-2x+2)) is 5/2 ln|x^2-2x+2|-arctan(x-1)+C

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integral of ((5x-6))/((x^2-2x+2))

integral

\frac{( 5 x - 6 )}{( x ^{2} - 2 x + 2 )}

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

Solution steps

\int \frac{5 x - 6}{x ^{2} - 2 x + 2} d x

Expand

\frac{5 x - 6}{x ^{2} - 2 x + 2} : \frac{5 x}{x ^{2} - 2 x + 2} - \frac{6}{x ^{2} - 2 x + 2}

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

What is the integral of ((5x-6))/((x^2-2x+2)) ?
The integral of ((5x-6))/((x^2-2x+2)) is 5/2 ln|x^2-2x+2|-arctan(x-1)+C