What is the integral of 1/(x^6+x^4) ?

The integral of 1/(x^6+x^4) is 1/x-1/(3x^3)+arctan(x)+C

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integral of 1/(x^6+x^4)

\int \frac{1}{x ^{6} + x ^{4}} d x

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

Solution steps

\int \frac{1}{x ^{6} + x ^{4}} d x

Take the partial fraction of

\frac{1}{x ^{6} + x ^{4}} : - \frac{1}{x ^{2}} + \frac{1}{x ^{4}} + \frac{1}{x ^{2} + 1}

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

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

\int \frac{1}{x ^{4}} d x = - \frac{1}{3 x ^{3}}

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

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

Simplify

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

Add a constant to the solution

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

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What is the integral of 1/(x^6+x^4) ?
The integral of 1/(x^6+x^4) is 1/x-1/(3x^3)+arctan(x)+C