What is the integral of 1/(x^8(1+x^2)) ?

The integral of 1/(x^8(1+x^2)) is 1/x-1/(3x^3)+1/(5x^5)-1/(7x^7)+arctan(x)+C

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integral of 1/(x^8(1+x^2))

\int \frac{1}{x ^{8} ( 1 + x ^{2} )} d x

\frac{1}{x} - \frac{1}{3 x ^{3}} + \frac{1}{5 x ^{5}} - \frac{1}{7 x ^{7}} + arctan (x) + C

Solution steps

\int \frac{1}{x ^{8} ( 1 + x ^{2} )} d x

Take the partial fraction of

\frac{1}{x ^{8} ( 1 + x ^{2} )} : - \frac{1}{x ^{2}} + \frac{1}{x ^{4}} - \frac{1}{x ^{6}} + \frac{1}{x ^{8}} + \frac{1}{x ^{2} + 1}

= \int - \frac{1}{x ^{2}} + \frac{1}{x ^{4}} - \frac{1}{x ^{6}} + \frac{1}{x ^{8}} + \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 ^{8}} 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 ^{8}} d x = - \frac{1}{7 x ^{7}}

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

= - (- \frac{1}{x}) - \frac{1}{3 x ^{3}} - (- \frac{1}{5 x ^{5}}) - \frac{1}{7 x ^{7}} + arctan (x)

Simplify

= \frac{1}{x} - \frac{1}{3 x ^{3}} + \frac{1}{5 x ^{5}} - \frac{1}{7 x ^{7}} + arctan (x)

Add a constant to the solution

= \frac{1}{x} - \frac{1}{3 x ^{3}} + \frac{1}{5 x ^{5}} - \frac{1}{7 x ^{7}} + arctan (x) + C

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What is the integral of 1/(x^8(1+x^2)) ?
The integral of 1/(x^8(1+x^2)) is 1/x-1/(3x^3)+1/(5x^5)-1/(7x^7)+arctan(x)+C