What is the integral of ((x-1))/(3x^2-4x+3) ?

The integral of ((x-1))/(3x^2-4x+3) is 1/6 ln|9x^2-12x+9|-1/(3sqrt(5))arctan((3x-2)/(sqrt(5)))+C

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integral of ((x-1))/(3x^2-4x+3)

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

\frac{1}{6} ln 9 x^{2} - 12 x + 9 - \frac{1}{3 5} arctan (\frac{3 x - 2}{5}) + C

Solution steps

\int \frac{x - 1}{3 x ^{2} - 4 x + 3} d x

Complete the square

3 x^{2} - 4 x + 3 : 3 (x - \frac{2}{3})^{2} + \frac{5}{3}

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

Apply u-substitution

= \int \frac{3 u - 1}{9 u ^{2} + 5} d u

Expand

\frac{3 u - 1}{9 u ^{2} + 5} : \frac{3 u}{9 u ^{2} + 5} - \frac{1}{9 u ^{2} + 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{3 u}{9 u ^{2} + 5} d u - \int \frac{1}{9 u ^{2} + 5} d u

\int \frac{3 u}{9 u ^{2} + 5} d u = \frac{1}{6} ln 9 u^{2} + 5

\int \frac{1}{9 u ^{2} + 5} d u = \frac{1}{3 5} arctan (\frac{3}{5} u)

= \frac{1}{6} ln 9 u^{2} + 5 - \frac{1}{3 5} arctan (\frac{3}{5} u)

Substitute back

u = x - \frac{2}{3}

= \frac{1}{6} ln 9 (x - \frac{2}{3})^{2} + 5 - \frac{1}{3 5} arctan (\frac{3}{5} (x - \frac{2}{3}))

Simplify

\frac{1}{6} ln 9 (x - \frac{2}{3})^{2} + 5 - \frac{1}{3 5} arctan (\frac{3}{5} (x - \frac{2}{3})) : \frac{1}{6} ln 9 x^{2} - 12 x + 9 - \frac{1}{3 5} arctan (\frac{3 x - 2}{5})

= \frac{1}{6} ln 9 x^{2} - 12 x + 9 - \frac{1}{3 5} arctan (\frac{3 x - 2}{5})

Add a constant to the solution

= \frac{1}{6} ln 9 x^{2} - 12 x + 9 - \frac{1}{3 5} arctan (\frac{3 x - 2}{5}) + C

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What is the integral of ((x-1))/(3x^2-4x+3) ?
The integral of ((x-1))/(3x^2-4x+3) is 1/6 ln|9x^2-12x+9|-1/(3sqrt(5))arctan((3x-2)/(sqrt(5)))+C