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

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

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

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

\frac{1}{2} arcsec (\frac{1}{2} x) + C

Solution steps

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

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

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

x^{2} = (x),

assuming

x \geq 0

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

Apply Trigonometric Substitution

: \int \frac{1}{2} d u

= \int \frac{1}{2} 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 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{1}{2} x)

= \frac{1}{2} \cdot 1 \cdot arcsec (\frac{1}{2} x)

Simplify

= \frac{1}{2} arcsec (\frac{1}{2} x)

Add a constant to the solution

= \frac{1}{2} arcsec (\frac{1}{2} x) + C

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