What is the integral of 1/(2x^2sqrt(x^2-9)) ?

The integral of 1/(2x^2sqrt(x^2-9)) is (sqrt(x^2-9))/(18x)+C

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integral of 1/(2x^2sqrt(x^2-9))

\int \frac{1}{2 x ^{2} x ^{2} - 9} d x

\frac{x ^{2} - 9}{18 x} + C

Solution steps

\int \frac{1}{2 x ^{2} x ^{2} - 9} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{x ^{2} x ^{2} - 9} d x

Apply Trigonometric Substitution

: \int \frac{1}{9 sec ( u )} d u

= \frac{1}{2} \cdot \int \frac{1}{9 sec ( u )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{9} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{2} \cdot \frac{1}{9} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{9} sin (u)

Substitute back

u = arcsec (\frac{1}{3} x)

= \frac{1}{2} \cdot \frac{1}{9} sin (arcsec (\frac{1}{3} x))

Simplify

\frac{1}{2} \cdot \frac{1}{9} sin (arcsec (\frac{1}{3} x)) : \frac{x ^{2} - 9}{18 x}

= \frac{x ^{2} - 9}{18 x}

Add a constant to the solution

= \frac{x ^{2} - 9}{18 x} + C

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y^{- 2}

x \to 1 lim (x - \frac{1}{ln ∣ x ∣})

6 x^{2} y^{^{'}} = y^{^{'}} + 8 x e^{- y}

What is the integral of 1/(2x^2sqrt(x^2-9)) ?
The integral of 1/(2x^2sqrt(x^2-9)) is (sqrt(x^2-9))/(18x)+C