Indefinite integral are written in the form of `int f(x) dx = F(x) +C`

where: `f(x)` as the integrand

`F(x)` as the anti-derivative function

`C` as the arbitrary constant known as constant of integration

For the given problem `int 1/(xsqrt(4x^2+9))...

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Indefinite integral are written in the form of `int f(x) dx = F(x) +C`

where: `f(x)` as the integrand

`F(x)` as the anti-derivative function

`C` as the arbitrary constant known as constant of integration

For the given problem `int 1/(xsqrt(4x^2+9)) dx` , it resembles one of the formula from integration table. We may apply the integral formula for rational function with roots as:

`int dx/(xsqrt(x^2+a^2))= -1/aln((a+sqrt(x^2+a^2))/x)+C` .

For easier comparison, we apply u-substitution by letting: `u^2 =4x^2` or `(2x)^2` then `u = 2x` or `u/2 =x` .

Note: The corresponding value of `a^2=9` or `3^2` then ` a=3` .

For the derivative of `u` , we get: `du = 2 dx` or ` (du)/2= dx` .

Plug-in the values on the integral problem, we get:

`int 1/(xsqrt(4x^2+9)) dx =int 1/((u/2)sqrt(u^2+9)) *(du)/2`

` =int 2/(usqrt(u^2+9)) *(du)/2`

`=int (du)/(usqrt(u^2+9))`

Applying the aforementioned integral formula where `a^2=9` and `a=3` , we get:

`int (du)/(usqrt(u^2+9)) =-1/3ln((3+sqrt(u^2+9))/u)+C`

Plug-in `u^2 =4x^2` and `u =2x` on `-1/3ln((3+sqrt(u^2+9))/u)+C` , we get the indefinite integral as:

`int 1/(xsqrt(4x^2+9)) dx=-1/3ln((3+sqrt(4x^2+9))/(2x))+C`