What is the integral of (x-3)/(x^4+9x^2) ?

The integral of (x-3)/(x^4+9x^2) is 1/9 ln(x)-1/18 ln|x^2+9|-3(-1/(9x)-1/27 arctan(x/3))+C

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integral of (x-3)/(x^4+9x^2)

\int \frac{x - 3}{x ^{4} + 9 x ^{2}} d x

\frac{1}{9} ln (x) - \frac{1}{18} ln x^{2} + 9 - 3 (- \frac{1}{9 x} - \frac{1}{27} arctan (\frac{x}{3})) + C

Solution steps

\int \frac{x - 3}{x ^{4} + 9 x ^{2}} d x

Expand

\frac{x - 3}{x ^{4} + 9 x ^{2}} : \frac{x}{x ^{4} + 9 x ^{2}} - \frac{3}{x ^{4} + 9 x ^{2}}

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{x}{x ^{4} + 9 x ^{2}} d x - \int \frac{3}{x ^{4} + 9 x ^{2}} d x

\int \frac{x}{x ^{4} + 9 x ^{2}} d x = \frac{1}{9} ln (x) - \frac{1}{18} ln x^{2} + 9

\int \frac{3}{x ^{4} + 9 x ^{2}} d x = 3 (- \frac{1}{9 x} - \frac{1}{27} arctan (\frac{x}{3}))

= \frac{1}{9} ln (x) - \frac{1}{18} ln x^{2} + 9 - 3 (- \frac{1}{9 x} - \frac{1}{27} arctan (\frac{x}{3}))

Add a constant to the solution

= \frac{1}{9} ln (x) - \frac{1}{18} ln x^{2} + 9 - 3 (- \frac{1}{9 x} - \frac{1}{27} arctan (\frac{x}{3})) + C

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What is the integral of (x-3)/(x^4+9x^2) ?
The integral of (x-3)/(x^4+9x^2) is 1/9 ln(x)-1/18 ln|x^2+9|-3(-1/(9x)-1/27 arctan(x/3))+C