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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

y^{^{'}} - y = 2 cos (5 t)

x \to 2 lim (- x^{3})

x \to 0 lim (\frac{1}{2 ^{x} - 1} - \frac{1}{x})

f (x) = \frac{1}{x ^{5}}, at x = 1

\int \frac{t ^{6}}{1 - t ^{14}} d t

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