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

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

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

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

\frac{1}{7} ln ∣ x - 2 ∣ + \frac{1}{7} (12 (\frac{1}{4} ln 4 x^{2} + 4 x + 4 - \frac{1}{2 3} arctan (\frac{2 x + 1}{3})) + \frac{22}{3} arctan (\frac{2 x + 1}{3})) + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} - 3}{( x - 2 ) ( x ^{2} + x + 1 )} : \frac{1}{7 ( x - 2 )} + \frac{6 x + 11}{7 ( x ^{2} + x + 1 )}

= \int \frac{1}{7 ( x - 2 )} + \frac{6 x + 11}{7 ( x ^{2} + 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}{7 ( x - 2 )} d x + \int \frac{6 x + 11}{7 ( x ^{2} + x + 1 )} d x

\int \frac{1}{7 ( x - 2 )} d x = \frac{1}{7} ln ∣ x - 2 ∣

\int \frac{6 x + 11}{7 ( x ^{2} + x + 1 )} d x = \frac{1}{7} (12 (\frac{1}{4} ln 4 x^{2} + 4 x + 4 - \frac{1}{2 3} arctan (\frac{2 x + 1}{3})) + \frac{22}{3} arctan (\frac{2 x + 1}{3}))

= \frac{1}{7} ln ∣ x - 2 ∣ + \frac{1}{7} (12 (\frac{1}{4} ln 4 x^{2} + 4 x + 4 - \frac{1}{2 3} arctan (\frac{2 x + 1}{3})) + \frac{22}{3} arctan (\frac{2 x + 1}{3}))

Add a constant to the solution

= \frac{1}{7} ln ∣ x - 2 ∣ + \frac{1}{7} (12 (\frac{1}{4} ln 4 x^{2} + 4 x + 4 - \frac{1}{2 3} arctan (\frac{2 x + 1}{3})) + \frac{22}{3} arctan (\frac{2 x + 1}{3})) + C

\int (x^{3} - 7 x) e^{5 x} d x

\int \frac{ln ( ln ( 5 x ) )}{x} d x

\int \frac{x}{x ^{2} + 5 x + 4} d x

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

\int \frac{x}{8 - 3 x ^{2}} d x

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