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`int 1/(xsqrt(4x^2+9)) dx` Find the indefinite integral

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`

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