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

The integral of 1/(sqrt(x^6-x^2)) is 1/2 arctan(sqrt(x^4-1))+C

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integral of 1/(sqrt(x^6-x^2))

\int \frac{1}{x ^{6} - x ^{2}} d x

\frac{1}{2} arctan (x^{4} - 1) + C

Solution steps

\int \frac{1}{x ^{6} - x ^{2}} d x

x^{6} - x^{2} = x x^{4} - 1

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

Apply u-substitution

: \int \frac{1}{2 ( u ^{2} + 1 )} d u

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= \frac{1}{2} arctan (u)

Substitute back

u = x^{4} - 1

= \frac{1}{2} arctan (x^{4} - 1)

Add a constant to the solution

= \frac{1}{2} arctan (x^{4} - 1) + C

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What is the integral of 1/(sqrt(x^6-x^2)) ?
The integral of 1/(sqrt(x^6-x^2)) is 1/2 arctan(sqrt(x^4-1))+C