What is the integral of 1/(x^7-4x^5+6x^3-4x) ?

The integral of 1/(x^7-4x^5+6x^3-4x) is -1/4 ln|x|+1/8 ln|x+sqrt(2)|+1/8 ln|x-sqrt(2)|+1/(4x^2)+C

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integral of 1/(x^7-4x^5+6x^3-4x)

\int \frac{1}{x ^{7} - 4 x ^{5} + 6 x ^{3} - 4 x} d x

- \frac{1}{4} ln ∣ x ∣ + \frac{1}{8} ln x + 2 + \frac{1}{8} ln x - 2 + \frac{1}{4 x ^{2}} + C

Solution steps

\int \frac{1}{x ^{7} - 4 x ^{5} + 6 x ^{3} - 4 x} d x

Take the partial fraction of

\frac{1}{x ^{7} - 4 x ^{5} + 6 x ^{3} - 4 x} : - \frac{1}{4 x} + \frac{1}{8 ( x + 2 )} + \frac{1}{8 ( x - 2 )} - \frac{x}{2 ( x ^{4} - 2 x ^{2} + 2 )}

= \int - \frac{1}{4 x} + \frac{1}{8 ( x + 2 )} + \frac{1}{8 ( x - 2 )} - \frac{x}{2 ( x ^{4} - 2 x ^{2} + 2 )} 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}{4 x} d x + \int \frac{1}{8 ( x + 2 )} d x + \int \frac{1}{8 ( x - 2 )} d x - \int \frac{x}{2 ( x ^{4} - 2 x ^{2} + 2 )} d x

\int \frac{1}{4 x} d x = \frac{1}{4} ln ∣ x ∣

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

\int \frac{1}{8 ( x - 2 )} d x = \frac{1}{8} ln x - 2

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

= - \frac{1}{4} ln ∣ x ∣ + \frac{1}{8} ln x + 2 + \frac{1}{8} ln x - 2 - (- \frac{1}{4 x ^{2}})

Simplify

= - \frac{1}{4} ln ∣ x ∣ + \frac{1}{8} ln x + 2 + \frac{1}{8} ln x - 2 + \frac{1}{4 x ^{2}}

Add a constant to the solution

= - \frac{1}{4} ln ∣ x ∣ + \frac{1}{8} ln x + 2 + \frac{1}{8} ln x - 2 + \frac{1}{4 x ^{2}} + C

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What is the integral of 1/(x^7-4x^5+6x^3-4x) ?
The integral of 1/(x^7-4x^5+6x^3-4x) is -1/4 ln|x|+1/8 ln|x+sqrt(2)|+1/8 ln|x-sqrt(2)|+1/(4x^2)+C