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

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

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

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

\frac{1}{7} (- ln 4 x^{2} - 4 x + 4 - \frac{2}{3} arctan (\frac{1}{3} (2 x - 1)) + 23 arctan (\frac{1}{3} (2 x - 1))) + \frac{2}{7} ln ∣ 3 x - 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x}{( x ^{2} - x + 1 ) ( 3 x - 2 )} : \frac{- 2 x + 3}{7 ( x ^{2} - x + 1 )} + \frac{6}{7 ( 3 x - 2 )}

= \int \frac{- 2 x + 3}{7 ( x ^{2} - x + 1 )} + \frac{6}{7 ( 3 x - 2 )} d x

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{- 2 x + 3}{7 ( x ^{2} - x + 1 )} d x + \int \frac{6}{7 ( 3 x - 2 )} d x

\int \frac{- 2 x + 3}{7 ( x ^{2} - x + 1 )} d x = \frac{1}{7} (- ln 4 x^{2} - 4 x + 4 - \frac{2}{3} arctan (\frac{1}{3} (2 x - 1)) + 23 arctan (\frac{1}{3} (2 x - 1)))

\int \frac{6}{7 ( 3 x - 2 )} d x = \frac{2}{7} ln ∣ 3 x - 2 ∣

= \frac{1}{7} (- ln 4 x^{2} - 4 x + 4 - \frac{2}{3} arctan (\frac{1}{3} (2 x - 1)) + 23 arctan (\frac{1}{3} (2 x - 1))) + \frac{2}{7} ln ∣ 3 x - 2 ∣

Add a constant to the solution

= \frac{1}{7} (- ln 4 x^{2} - 4 x + 4 - \frac{2}{3} arctan (\frac{1}{3} (2 x - 1)) + 23 arctan (\frac{1}{3} (2 x - 1))) + \frac{2}{7} ln ∣ 3 x - 2 ∣ + C

\int \frac{( 4 x + 13 )}{x ^{2} + 3 x - 10} d x

\int_{0}^{\frac{π}{2}} \frac{5 cos ( x )}{sin ( x )} d x

\frac{d y}{d t} = \frac{4 t ^{8}}{e + t ^{9}}

\frac{d}{d x} (\frac{- 2}{( x - 1 ) ^{2}})

y^{^{'}} = e^{5 x} - 9 x

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