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`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`

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