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

The integral of 4/(x^4sqrt(x^2-8)) is ((x^2+4)sqrt(x^2-8))/(24x^3)+C

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

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

\frac{( x ^{2} + 4 ) x ^{2} - 8}{24 x ^{3}} + C

Solution steps

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

Take the constant out:

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

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

Apply Trigonometric Substitution

= 4 \cdot \int \frac{1}{64 sec ^{3} ( u )} d u

Take the constant out:

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

= 4 \cdot \frac{1}{64} \cdot \int \frac{1}{sec ^{3} ( u )} d u

Rewrite using trig identities

= 4 \cdot \frac{1}{64} \cdot \int (1 - sin^{2} (u)) cos (u) d u

Apply u-substitution

= 4 \cdot \frac{1}{64} \cdot \int 1 - v^{2} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 \cdot \frac{1}{64} (\int 1 d v - \int v^{2} d v)

\int 1 d v = v

\int v^{2} d v = \frac{v ^{3}}{3}

= 4 \cdot \frac{1}{64} (v - \frac{v ^{3}}{3})

Substitute back

= 4 \cdot \frac{1}{64} sin (arcsec (\frac{1}{2 2} x)) - \frac{sin ^{3} ( arcsec ( \frac{1}{2 2} x ) )}{3}

Simplify

4 \cdot \frac{1}{64} sin (arcsec (\frac{1}{2 2} x)) - \frac{sin ^{3} ( arcsec ( \frac{1}{2 2} x ) )}{3} : \frac{( x ^{2} + 4 ) x ^{2} - 8}{24 x ^{3}}

= \frac{( x ^{2} + 4 ) x ^{2} - 8}{24 x ^{3}}

Add a constant to the solution

= \frac{( x ^{2} + 4 ) x ^{2} - 8}{24 x ^{3}} + C

What is the integral of 4/(x^4sqrt(x^2-8)) ?
The integral of 4/(x^4sqrt(x^2-8)) is ((x^2+4)sqrt(x^2-8))/(24x^3)+C