What is the integral of sqrt((x^4-8)/(x^{14))} ?

The integral of sqrt((x^4-8)/(x^{14))} is ((x^4-8)^{3/2})/(48x^6)+C

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integral of sqrt((x^4-8)/(x^{14))}

\int \frac{x ^{4} - 8}{x ^{14}} d x

\frac{( x ^{4} - 8 ) ^{\frac{3}{2}}}{48 x ^{6}} + C

Solution steps

\int \frac{x ^{4} - 8}{x ^{14}} d x

Apply u-substitution

: \int \frac{u ^{2} - 8}{2 u ^{4}} d u

= \int \frac{u ^{2} - 8}{2 u ^{4}} 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 \frac{u ^{2} - 8}{u ^{4}} d u

Apply Trigonometric Substitution

: \int \frac{tan ^{2} ( v )}{8 sec ^{3} ( v )} d v

= \frac{1}{2} \cdot \int \frac{tan ^{2} ( v )}{8 sec ^{3} ( v )} d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{8} \cdot \int \frac{tan ^{2} ( v )}{sec ^{3} ( v )} d v

Rewrite using trig identities

= \frac{1}{2} \cdot \frac{1}{8} \cdot \int sin^{2} (v) cos (v) d v

Apply u-substitution

: \int w^{2} d w

= \frac{1}{2} \cdot \frac{1}{8} \cdot \int w^{2} d w

Apply the Power Rule

: \frac{w ^{3}}{3}

= \frac{1}{2} \cdot \frac{1}{8} \cdot \frac{w ^{3}}{3}

Substitute back

= \frac{1}{2} \cdot \frac{1}{8} \cdot \frac{sin ^{3} ( arcsec ( \frac{1}{2 2} x ^{2} ) )}{3}

Simplify

\frac{1}{2} \cdot \frac{1}{8} \cdot \frac{sin ^{3} ( arcsec ( \frac{1}{2 2} x ^{2} ) )}{3} : \frac{( x ^{4} - 8 ) ^{\frac{3}{2}}}{48 x ^{6}}

= \frac{( x ^{4} - 8 ) ^{\frac{3}{2}}}{48 x ^{6}}

Add a constant to the solution

= \frac{( x ^{4} - 8 ) ^{\frac{3}{2}}}{48 x ^{6}} + C

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What is the integral of sqrt((x^4-8)/(x^{14))} ?
The integral of sqrt((x^4-8)/(x^{14))} is ((x^4-8)^{3/2})/(48x^6)+C