What is the integral of 2/((x-1/x)sqrt(x^4-2x^2-8)) ?

The integral of 2/((x-1/x)sqrt(x^4-2x^2-8)) is 1/3 arcsec(1/3 (x^2-1))+C

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integral of 2/((x-1/x)sqrt(x^4-2x^2-8))

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

\frac{1}{3} arcsec (\frac{1}{3} (x^{2} - 1)) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{2 ( u - 1 ) u ^{2} - 2 u - 8} d u

= 2 \cdot \int \frac{1}{2 ( u - 1 ) u ^{2} - 2 u - 8} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{( u - 1 ) u ^{2} - 2 u - 8} d u

Complete the square

u^{2} - 2 u - 8 : (u - 1)^{2} - 9

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{( u - 1 ) ( u - 1 ) ^{2} - 9} d u

Apply u-substitution

: \int \frac{1}{v v ^{2} - 9} d v

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{v v ^{2} - 9} d v

Apply Trigonometric Substitution

: \int \frac{1}{3} d w

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{3} d w

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \int 1 d w

Integral of a constant:

\int a d x = a x

= 2 \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot 1 \cdot w

Substitute back

= 2 \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot 1 \cdot arcsec (\frac{1}{3} (x^{2} - 1))

Simplify

2 \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot 1 \cdot arcsec (\frac{1}{3} (x^{2} - 1)) : \frac{1}{3} arcsec (\frac{1}{3} (x^{2} - 1))

= \frac{1}{3} arcsec (\frac{1}{3} (x^{2} - 1))

Add a constant to the solution

= \frac{1}{3} arcsec (\frac{1}{3} (x^{2} - 1)) + C

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What is the integral of 2/((x-1/x)sqrt(x^4-2x^2-8)) ?
The integral of 2/((x-1/x)sqrt(x^4-2x^2-8)) is 1/3 arcsec(1/3 (x^2-1))+C