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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Add a constant to the solution

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

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