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

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

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

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

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

Solution steps

\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}

= \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

= \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)

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

Add a constant to the solution

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

x^{2} e^{\frac{1}{x}}

\frac{\partial}{\partial x} (t^{2} x - t)

\int \frac{x ^{2}}{x ^{4} - 20 x ^{2} - 125} d x

\frac{d}{d x} (5 \cdot 4^{x})

\frac{d}{d x} (lo g_{4} (x) (2^{3 x} + e^{x^{4}})^{- 2})

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