What is the integral of sqrt((x^4-2)/(x^{14))} ?

The integral of sqrt((x^4-2)/(x^{14))} is 1/2 (-(sqrt(x^4-2))/(3x^6)+(sqrt(x^4-2))/(6x^2))+C

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integral of sqrt((x^4-2)/(x^{14))}

\int \frac{x ^{4} - 2}{x ^{14}} d x

\frac{1}{2} (- \frac{x ^{4} - 2}{3 x ^{6}} + \frac{x ^{4} - 2}{6 x ^{2}}) + C

Solution steps

\int \frac{x ^{4} - 2}{x ^{14}} d x

Apply u-substitution

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

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

Apply Integration By Parts

: - \frac{u ^{2} - 2}{3 u ^{3}} - \int - \frac{1}{3 u ^{2} u ^{2} - 2} d u

= \frac{1}{2} (- \frac{u ^{2} - 2}{3 u ^{3}} - \int - \frac{1}{3 u ^{2} u ^{2} - 2} d u)

\int - \frac{1}{3 u ^{2} u ^{2} - 2} d u = - \frac{u ^{2} - 2}{6 u}

= \frac{1}{2} (- \frac{u ^{2} - 2}{3 u ^{3}} - (- \frac{u ^{2} - 2}{6 u}))

Substitute back

u = x^{2}

= \frac{1}{2} - \frac{( x ^{2} ) ^{2} - 2}{3 ( x ^{2} ) ^{3}} - - \frac{( x ^{2} ) ^{2} - 2}{6 x ^{2}}

Simplify

\frac{1}{2} - \frac{( x ^{2} ) ^{2} - 2}{3 ( x ^{2} ) ^{3}} - - \frac{( x ^{2} ) ^{2} - 2}{6 x ^{2}} : \frac{1}{2} (- \frac{x ^{4} - 2}{3 x ^{6}} + \frac{x ^{4} - 2}{6 x ^{2}})

= \frac{1}{2} (- \frac{x ^{4} - 2}{3 x ^{6}} + \frac{x ^{4} - 2}{6 x ^{2}})

Add a constant to the solution

= \frac{1}{2} (- \frac{x ^{4} - 2}{3 x ^{6}} + \frac{x ^{4} - 2}{6 x ^{2}}) + C

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What is the integral of sqrt((x^4-2)/(x^{14))} ?
The integral of sqrt((x^4-2)/(x^{14))} is 1/2 (-(sqrt(x^4-2))/(3x^6)+(sqrt(x^4-2))/(6x^2))+C