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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

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

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

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

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

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

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

Add a constant to the solution

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

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