Put `u=x^2+9`

Then `du = 2x\ dx`

The question has just one `x\ dx` so we divide both sides by 2:

`(du)/2=x\ dx`

Now

`intx/sqrt(x^2+9)dx`

`=1/2int(du)/sqrtu`

`=1/2intu^(-1//2)du`

`=1/2(u^(1//2))/(1//2)+K`

`=u^(1//2)+K`

`=sqrt(x^2+9)+K`