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

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

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

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

ln ∣ x - 2 ∣ + ln 4 x^{2} + 12 x + 4 + \frac{2}{5} (ln \frac{1}{5} (2 x + 3) + 1 - ln \frac{1}{5} (2 x + 3) - 1) + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x - 2} d x = ln ∣ x - 2 ∣

\int \frac{2 x + 1}{x ^{2} + 3 x + 1} d x = ln 4 x^{2} + 12 x + 4 + \frac{2}{5} (ln \frac{1}{5} (2 x + 3) + 1 - ln \frac{1}{5} (2 x + 3) - 1)

= ln ∣ x - 2 ∣ + ln 4 x^{2} + 12 x + 4 + \frac{2}{5} (ln \frac{1}{5} (2 x + 3) + 1 - ln \frac{1}{5} (2 x + 3) - 1)

Add a constant to the solution

= ln ∣ x - 2 ∣ + ln 4 x^{2} + 12 x + 4 + \frac{2}{5} (ln \frac{1}{5} (2 x + 3) + 1 - ln \frac{1}{5} (2 x + 3) - 1) + C

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