What is the integral of (2x^2-7x+5)/((x^2+4)(x^2+7)) ?

The integral of (2x^2-7x+5)/((x^2+4)(x^2+7)) is

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integral of (2x^2-7x+5)/((x^2+4)(x^2+7))

\int \frac{2 x ^{2} - 7 x + 5}{( x ^{2} + 4 ) ( x ^{2} + 7 )} d x

\frac{1}{6} (- 7 ln \frac{x ^{2}}{4} + 1 - 3 arctan (\frac{x}{2})) + \frac{1}{3} (\frac{7}{2} ln x^{2} + 7 + \frac{9}{7} arctan (\frac{x}{7})) + C

Solution steps

\int \frac{2 x ^{2} - 7 x + 5}{( x ^{2} + 4 ) ( x ^{2} + 7 )} d x

Take the partial fraction of

\frac{2 x ^{2} - 7 x + 5}{( x ^{2} + 4 ) ( x ^{2} + 7 )} : \frac{- 7 x - 3}{3 ( x ^{2} + 4 )} + \frac{7 x + 9}{3 ( x ^{2} + 7 )}

= \int \frac{- 7 x - 3}{3 ( x ^{2} + 4 )} + \frac{7 x + 9}{3 ( x ^{2} + 7 )} 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{- 7 x - 3}{3 ( x ^{2} + 4 )} d x + \int \frac{7 x + 9}{3 ( x ^{2} + 7 )} d x

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

\int \frac{7 x + 9}{3 ( x ^{2} + 7 )} d x = \frac{1}{3} (\frac{7}{2} ln x^{2} + 7 + \frac{9}{7} arctan (\frac{x}{7}))

= \frac{1}{6} (- 7 ln \frac{x ^{2}}{4} + 1 - 3 arctan (\frac{x}{2})) + \frac{1}{3} (\frac{7}{2} ln x^{2} + 7 + \frac{9}{7} arctan (\frac{x}{7}))

Add a constant to the solution

= \frac{1}{6} (- 7 ln \frac{x ^{2}}{4} + 1 - 3 arctan (\frac{x}{2})) + \frac{1}{3} (\frac{7}{2} ln x^{2} + 7 + \frac{9}{7} arctan (\frac{x}{7})) + C

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What is the integral of (2x^2-7x+5)/((x^2+4)(x^2+7)) ?
The integral of (2x^2-7x+5)/((x^2+4)(x^2+7)) is