int 1/(x^2sqrt(2+9x^2)) dx Use integration tables to find the indefinite integral.

Recall that indefinite integral follows int f(x) dx = F(x) +C where:
f(x) as the integrand function
F(x) as the antiderivative of f(x)
C as the constant of integration.
The given integral problem: int 1/(x^2sqrt(2+9x^2)) dx resembles one of the formulas from the integration table. We follow the integral formula for function with roots as:
int 1/(u^2sqrt(a^2+u^2))du =- sqrt(a^2+u^2)/(a^2u) +C
We apply u-substitution by letting: u^2 = 9x^2 or (3x)^2 then u = 3x  or x=u/3 .
For the derivative of u, we get: du = 3 dx or (du)/3 = dx .
Note: The corresponding value of a^2=2 .
Plug-in the values of u = 3x , x=u/3 and (du)/3 = dx , we get:
int 1/(x^2sqrt(2+9x^2))dx=int 1/((u/3)^2sqrt(2+u^2))* (du)/3
                               =int 1/(u^2/9*sqrt(2+u^2))* (du)/3
                               =int 9/(u^2sqrt(2+u^2))* (du)/3
                               =int (3 du)/(u^2sqrt(2+u^2))
Apply the basic integration property: int c*f(x) dx = c int f(x) dx .
int (3 du)/(u^2sqrt(2+u^2))=3int (du)/(u^2sqrt(2+u^2))
Apply the aforementioned integral formula with a^2 =2 , we get:
3int (du)/(u^2sqrt(2+u^2)) = 3*[- sqrt(2+u^2)/(2u)] +C
                          =-(3 sqrt(2+u^2))/(2u) +C
Plug-in u =3x on -(3 sqrt(2+u^2))/(2u) +C , we get the indefinite integral as:
int 1/(x^2sqrt(2+9x^2)) dx =-(3 sqrt(2+(3x)^2))/(2*(3x)) +C
                               =-(3 sqrt(2+9x^2))/(6x) +C
                               =- sqrt(2+9x^2)/(2x) +C