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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

Rewrite using trig identities

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

Use the common integral:

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

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (4 x - 2 y + 3)

n = 1 \sum \infty \frac{x ^{n}}{n + 2}

\frac{d}{d x} ((9 x^{2} - 15 x) e^{x})

\int \frac{1}{9 sin ^{2} ( x )} d x

x \to 0 lim (\frac{tan ( 4 x )}{tan ( 9 x )})

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