What is the integral of sqrt(1/(x^4)-1/(x^2)) ?

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

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integral of sqrt(1/(x^4)-1/(x^2))

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

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

Solution steps

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

Simplify

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

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

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

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