What is the integral of 1/(x^6(1+x^2)) ?

The integral of 1/(x^6(1+x^2)) is -1/x+1/(3x^3)-1/(5x^5)-arctan(x)+C

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integral of 1/(x^6(1+x^2))

\int \frac{1}{x ^{6} ( 1 + x ^{2} )} d x

- \frac{1}{x} + \frac{1}{3 x ^{3}} - \frac{1}{5 x ^{5}} - arctan (x) + C

Solution steps

\int \frac{1}{x ^{6} ( 1 + x ^{2} )} d x

Take the partial fraction of

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

= \int \frac{1}{x ^{2}} - \frac{1}{x ^{4}} + \frac{1}{x ^{6}} - \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 ^{6}} 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 ^{6}} d x = - \frac{1}{5 x ^{5}}

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

= - \frac{1}{x} - (- \frac{1}{3 x ^{3}}) - \frac{1}{5 x ^{5}} - arctan (x)

Simplify

= - \frac{1}{x} + \frac{1}{3 x ^{3}} - \frac{1}{5 x ^{5}} - arctan (x)

Add a constant to the solution

= - \frac{1}{x} + \frac{1}{3 x ^{3}} - \frac{1}{5 x ^{5}} - arctan (x) + C

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What is the integral of 1/(x^6(1+x^2)) ?
The integral of 1/(x^6(1+x^2)) is -1/x+1/(3x^3)-1/(5x^5)-arctan(x)+C