What is the integral of (3x^2+x-2)/(x^4+8x^2+16) ?

The integral of (3x^2+x-2)/(x^4+8x^2+16) is

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integral of (3x^2+x-2)/(x^4+8x^2+16)

\int \frac{3 x ^{2} + x - 2}{x ^{4} + 8 x ^{2} + 16} d x

\frac{3 ( x ^{2} arctan ( \frac{x}{2} ) - 2 x + 4 arctan ( \frac{x}{2} ) )}{4 ( 4 + x ^{2} )} - \frac{1}{2 ( x ^{2} + 4 )} - \frac{1}{8} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2}))) + C

Solution steps

\int \frac{3 x ^{2} + x - 2}{x ^{4} + 8 x ^{2} + 16} d x

Expand

\frac{3 x ^{2} + x - 2}{x ^{4} + 8 x ^{2} + 16} : \frac{3 x ^{2}}{x ^{4} + 8 x ^{2} + 16} + \frac{x}{x ^{4} + 8 x ^{2} + 16} - \frac{2}{x ^{4} + 8 x ^{2} + 16}

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{3 x ^{2}}{x ^{4} + 8 x ^{2} + 16} d x + \int \frac{x}{x ^{4} + 8 x ^{2} + 16} d x - \int \frac{2}{x ^{4} + 8 x ^{2} + 16} d x

\int \frac{3 x ^{2}}{x ^{4} + 8 x ^{2} + 16} d x = \frac{3 ( x ^{2} arctan ( \frac{x}{2} ) - 2 x + 4 arctan ( \frac{x}{2} ) )}{4 ( 4 + x ^{2} )}

\int \frac{x}{x ^{4} + 8 x ^{2} + 16} d x = - \frac{1}{2 ( x ^{2} + 4 )}

\int \frac{2}{x ^{4} + 8 x ^{2} + 16} d x = \frac{1}{8} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2})))

= \frac{3 ( x ^{2} arctan ( \frac{x}{2} ) - 2 x + 4 arctan ( \frac{x}{2} ) )}{4 ( 4 + x ^{2} )} - \frac{1}{2 ( x ^{2} + 4 )} - \frac{1}{8} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2})))

Add a constant to the solution

= \frac{3 ( x ^{2} arctan ( \frac{x}{2} ) - 2 x + 4 arctan ( \frac{x}{2} ) )}{4 ( 4 + x ^{2} )} - \frac{1}{2 ( x ^{2} + 4 )} - \frac{1}{8} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2}))) + C

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What is the integral of (3x^2+x-2)/(x^4+8x^2+16) ?
The integral of (3x^2+x-2)/(x^4+8x^2+16) is