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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{1}{4 sec ^{3} ( u )} d u

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{cos ( 3 x ^{3} )}{tan ( 3 x ^{2} + 2 x )})

- \frac{3}{2} x^{2} + 4 x + 4

\int x^{- \frac{1}{5}} d x

y = - \frac{7}{( t + 5 ) ^{2}}

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