What is the integral of 1/(x^4sqrt(x^2-3)) ?

The integral of 1/(x^4sqrt(x^2-3)) is 1/9 ((sqrt(x^2-3))/x-((x^2-3)^{3/2})/(3x^3))+C

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integral of 1/(x^4sqrt(x^2-3))

\int \frac{1}{x ^{4} x ^{2} - 3} d x

\frac{1}{9} \frac{x ^{2} - 3}{x} - \frac{( x ^{2} - 3 ) ^{\frac{3}{2}}}{3 x ^{3}} + C

Solution steps

\int \frac{1}{x ^{4} x ^{2} - 3} d x

Apply Trigonometric Substitution

: \int \frac{1}{9 sec ^{3} ( u )} d u

= \int \frac{1}{9 sec ^{3} ( u )} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{1}{sec ^{3} ( u )} d u

Rewrite using trig identities

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

Apply u-substitution

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

= \frac{1}{9} \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

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

\int 1 d v = v

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

= \frac{1}{9} (v - \frac{v ^{3}}{3})

Substitute back

= \frac{1}{9} sin (arcsec (\frac{1}{3} x)) - \frac{sin ^{3} ( arcsec ( \frac{1}{3} x ) )}{3}

Simplify

\frac{1}{9} sin (arcsec (\frac{1}{3} x)) - \frac{sin ^{3} ( arcsec ( \frac{1}{3} x ) )}{3} : \frac{1}{9} \frac{x ^{2} - 3}{x} - \frac{( x ^{2} - 3 ) ^{\frac{3}{2}}}{3 x ^{3}}

= \frac{1}{9} \frac{x ^{2} - 3}{x} - \frac{( x ^{2} - 3 ) ^{\frac{3}{2}}}{3 x ^{3}}

Add a constant to the solution

= \frac{1}{9} \frac{x ^{2} - 3}{x} - \frac{( x ^{2} - 3 ) ^{\frac{3}{2}}}{3 x ^{3}} + C

\frac{d}{d u} (arctan (\frac{u}{v}))

\frac{d}{d x} (2 cos (4 x))

(x^{2} + 1) \cdot e^{x - 1}, 1

\int 11 tan^{4} (x) sec^{6} (x) d x

\int e^{x} e^{2 x} - 1 d x

What is the integral of 1/(x^4sqrt(x^2-3)) ?
The integral of 1/(x^4sqrt(x^2-3)) is 1/9 ((sqrt(x^2-3))/x-((x^2-3)^{3/2})/(3x^3))+C