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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

= arctan (x) - (- \frac{1}{2} ln x^{2} + 1 + ln (x)) - 2 (- \frac{1}{x} - arctan (x))

Simplify

arctan (x) - (- \frac{1}{2} ln x^{2} + 1 + ln (x)) - 2 (- \frac{1}{x} - arctan (x)) : \frac{1}{2} ln x^{2} + 1 + 3 arctan (x) - ln (x) + \frac{2}{x}

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

Add a constant to the solution

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

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