What is the integral of 1/((x^2+3x+5)(x^2-3x+4)) ?

The integral of 1/((x^2+3x+5)(x^2-3x+4)) is

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integral of 1/((x^2+3x+5)(x^2-3x+4))

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

\frac{3}{163} ln 4 x^{2} + 12 x + 20 + \frac{16}{163 11} arctan (\frac{2 x + 3}{11}) - \frac{3}{163} ln 4 x^{2} - 12 x + 16 + \frac{20}{163 7} arctan (\frac{2 x - 3}{7}) + C

Solution steps

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

Take the partial fraction of

\frac{1}{( x ^{2} + 3 x + 5 ) ( x ^{2} - 3 x + 4 )} : \frac{6 x + 17}{163 ( x ^{2} + 3 x + 5 )} + \frac{- 6 x + 19}{163 ( x ^{2} - 3 x + 4 )}

= \int \frac{6 x + 17}{163 ( x ^{2} + 3 x + 5 )} + \frac{- 6 x + 19}{163 ( x ^{2} - 3 x + 4 )} 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{6 x + 17}{163 ( x ^{2} + 3 x + 5 )} d x + \int \frac{- 6 x + 19}{163 ( x ^{2} - 3 x + 4 )} d x

\int \frac{6 x + 17}{163 ( x ^{2} + 3 x + 5 )} d x = \frac{3}{163} ln 4 x^{2} + 12 x + 20 + \frac{16}{163 11} arctan (\frac{2 x + 3}{11})

\int \frac{- 6 x + 19}{163 ( x ^{2} - 3 x + 4 )} d x = - \frac{3}{163} ln 4 x^{2} - 12 x + 16 + \frac{20}{163 7} arctan (\frac{2 x - 3}{7})

= \frac{3}{163} ln 4 x^{2} + 12 x + 20 + \frac{16}{163 11} arctan (\frac{2 x + 3}{11}) - \frac{3}{163} ln 4 x^{2} - 12 x + 16 + \frac{20}{163 7} arctan (\frac{2 x - 3}{7})

Add a constant to the solution

= \frac{3}{163} ln 4 x^{2} + 12 x + 20 + \frac{16}{163 11} arctan (\frac{2 x + 3}{11}) - \frac{3}{163} ln 4 x^{2} - 12 x + 16 + \frac{20}{163 7} arctan (\frac{2 x - 3}{7}) + C

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What is the integral of 1/((x^2+3x+5)(x^2-3x+4)) ?
The integral of 1/((x^2+3x+5)(x^2-3x+4)) is