What is the integral of sqrt((x^2-3)/(x^8)) ?

The integral of sqrt((x^2-3)/(x^8)) is ((x^2-3)^{3/2})/(9x^3)+C

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integral of sqrt((x^2-3)/(x^8))

\int \frac{x ^{2} - 3}{x ^{8}} d x

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

Solution steps

\int \frac{x ^{2} - 3}{x ^{8}} d x

Simplify

\frac{x ^{2} - 3}{x ^{8}} : \frac{x ^{2} - 3}{x ^{8}}

= \int \frac{x ^{2} - 3}{x ^{8}} d x

x^{8} = x^{4},

assuming

x \geq 0

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

Apply Trigonometric Substitution

: \int \frac{tan ^{2} ( u )}{3 sec ^{3} ( u )} d u

= \int \frac{tan ^{2} ( u )}{3 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}{3} \cdot \int \frac{tan ^{2} ( u )}{sec ^{3} ( u )} d u

Rewrite using trig identities

= \frac{1}{3} \cdot \int sin^{2} (u) cos (u) d u

Apply u-substitution

: \int v^{2} d v

= \frac{1}{3} \cdot \int v^{2} d v

Apply the Power Rule

: \frac{v ^{3}}{3}

= \frac{1}{3} \cdot \frac{v ^{3}}{3}

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int 2 x^{2} (- x^{3} + 3)^{5} d x

\frac{\partial}{\partial t} (3 arcsin (t))

\frac{d y}{d x} = - (1 + y)^{4}

\frac{d}{d x} (\frac{1}{2} x + 4)

\frac{\partial}{\partial y} (2 x y + 3 y)

What is the integral of sqrt((x^2-3)/(x^8)) ?
The integral of sqrt((x^2-3)/(x^8)) is ((x^2-3)^{3/2})/(9x^3)+C