What is the integral of 1/(x^4+3x^2+2) ?

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

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integral of 1/(x^4+3x^2+2)

\int \frac{1}{x ^{4} + 3 x ^{2} + 2} d x

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

Solution steps

\int \frac{1}{x ^{4} + 3 x ^{2} + 2} d x

Take the partial fraction of

\frac{1}{x ^{4} + 3 x ^{2} + 2} : \frac{1}{x ^{2} + 1} - \frac{1}{x ^{2} + 2}

= \int \frac{1}{x ^{2} + 1} - \frac{1}{x ^{2} + 2} 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{1}{x ^{2} + 1} d x - \int \frac{1}{x ^{2} + 2} d x

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

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

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

Add a constant to the solution

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

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