What is the integral of 1/(x^2sqrt(2x^2-32)) ?

The integral of 1/(x^2sqrt(2x^2-32)) is (sqrt(x^2-16))/(16sqrt(2)x)+C

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integral of 1/(x^2sqrt(2x^2-32))

\int \frac{1}{x ^{2} 2 x ^{2} - 32} d x

\frac{x ^{2} - 16}{16 2 x} + C

Solution steps

\int \frac{1}{x ^{2} 2 x ^{2} - 32} d x

Apply Trigonometric Substitution

: \int \frac{1}{16 2 sec ( u )} d u

= \int \frac{1}{16 2 sec ( u )} d u

Take the constant out:

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

= \frac{1}{16 2} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{16 2} \cdot \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= \frac{1}{16 2} sin (u)

Substitute back

u = arcsec (\frac{2}{4 2} x)

= \frac{1}{16 2} sin (arcsec (\frac{2}{4 2} x))

Simplify

\frac{1}{16 2} sin (arcsec (\frac{2}{4 2} x)) : \frac{x ^{2} - 16}{16 2 x}

= \frac{x ^{2} - 16}{16 2 x}

Add a constant to the solution

= \frac{x ^{2} - 16}{16 2 x} + C

\int sin^{4} (2 x) \cdot cos (2 x) d x

y = \frac{1}{( 1 + c _{1} e ^{- x} )}

\frac{\partial}{\partial x} (e^{x t})

\frac{d}{d x} (\frac{1}{x ^{\frac{1}{3}}})

\int z^{2} sin (x) d x

What is the integral of 1/(x^2sqrt(2x^2-32)) ?
The integral of 1/(x^2sqrt(2x^2-32)) is (sqrt(x^2-16))/(16sqrt(2)x)+C