What is the integral of 1/(x^4+13x^2+36) ?

The integral of 1/(x^4+13x^2+36) is

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integral of 1/(x^4+13x^2+36)

\int \frac{1}{x ^{4} + 13 x ^{2} + 36} d x

\frac{x}{( x ^{2} + 4 ) ( x ^{2} + 9 )} + \frac{1}{10} arctan (\frac{x}{2}) - \frac{1}{20} sin (2 arctan (\frac{x}{2})) - \frac{1}{15} arctan (\frac{x}{3}) + \frac{1}{30} sin (2 arctan (\frac{x}{3})) + C

Solution steps

\int \frac{1}{x ^{4} + 13 x ^{2} + 36} d x

Factor

x^{4} + 13 x^{2} + 36 : (x^{2} + 4) (x^{2} + 9)

= \int \frac{1}{( x ^{2} + 4 ) ( x ^{2} + 9 )} d x

Apply Integration By Parts

= \frac{x}{( x ^{2} + 4 ) ( x ^{2} + 9 )} - \int - \frac{x ( 4 x ^{3} + 26 x )}{( x ^{2} + 4 ) ^{2} ( x ^{2} + 9 ) ^{2}} d x

\int - \frac{x ( 4 x ^{3} + 26 x )}{( x ^{2} + 4 ) ^{2} ( x ^{2} + 9 ) ^{2}} d x = - \frac{1}{10} arctan (\frac{x}{2}) + \frac{1}{20} sin (2 arctan (\frac{x}{2})) + \frac{1}{15} arctan (\frac{x}{3}) - \frac{1}{30} sin (2 arctan (\frac{x}{3}))

= \frac{x}{( x ^{2} + 4 ) ( x ^{2} + 9 )} - (- \frac{1}{10} arctan (\frac{x}{2}) + \frac{1}{20} sin (2 arctan (\frac{x}{2})) + \frac{1}{15} arctan (\frac{x}{3}) - \frac{1}{30} sin (2 arctan (\frac{x}{3})))

Simplify

= \frac{x}{( x ^{2} + 4 ) ( x ^{2} + 9 )} + \frac{1}{10} arctan (\frac{x}{2}) - \frac{1}{20} sin (2 arctan (\frac{x}{2})) - \frac{1}{15} arctan (\frac{x}{3}) + \frac{1}{30} sin (2 arctan (\frac{x}{3}))

Add a constant to the solution

= \frac{x}{( x ^{2} + 4 ) ( x ^{2} + 9 )} + \frac{1}{10} arctan (\frac{x}{2}) - \frac{1}{20} sin (2 arctan (\frac{x}{2})) - \frac{1}{15} arctan (\frac{x}{3}) + \frac{1}{30} sin (2 arctan (\frac{x}{3})) + C