What is the integral of (x-1)/(9x^2-18x+17) ?

The integral of (x-1)/(9x^2-18x+17) is 1/18 ln|9x^2-18x+17|+C

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integral of (x-1)/(9x^2-18x+17)

\int \frac{x - 1}{9 x ^{2} - 18 x + 17} d x

\frac{1}{18} ln 9 x^{2} - 18 x + 17 + C

Solution steps

\int \frac{x - 1}{9 x ^{2} - 18 x + 17} d x

Expand

\frac{x - 1}{9 x ^{2} - 18 x + 17} : \frac{x}{9 x ^{2} - 18 x + 17} - \frac{1}{9 x ^{2} - 18 x + 17}

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}{9 x ^{2} - 18 x + 17} d x - \int \frac{1}{9 x ^{2} - 18 x + 17} d x

\int \frac{x}{9 x ^{2} - 18 x + 17} d x = \frac{1}{18} ln 9 x^{2} - 18 x + 17 + \frac{1}{6 2} arctan (\frac{3}{2 2} (x - 1))

\int \frac{1}{9 x ^{2} - 18 x + 17} d x = \frac{1}{6 2} arctan (\frac{3}{2 2} (x - 1))

= \frac{1}{18} ln 9 x^{2} - 18 x + 17 + \frac{1}{6 2} arctan (\frac{3}{2 2} (x - 1)) - \frac{1}{6 2} arctan (\frac{3}{2 2} (x - 1))

Add similar elements:

\frac{1}{6 2} arctan (\frac{3}{2 2} (x - 1)) - \frac{1}{6 2} arctan (\frac{3}{2 2} (x - 1)) = 0

= \frac{1}{18} ln 9 x^{2} - 18 x + 17

Add a constant to the solution

= \frac{1}{18} ln 9 x^{2} - 18 x + 17 + C

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What is the integral of (x-1)/(9x^2-18x+17) ?
The integral of (x-1)/(9x^2-18x+17) is 1/18 ln|9x^2-18x+17|+C