So i need to solve:
The integral of 1/ Sqrt( 9x^2 - 16) dx
If the 9 wasnt infront of x^2, i could easily do it, letting x = 4cosh(u) and so on, but the 9 confuses me.
Am i right in thinking i take a 9 out, to leave :
1/ sqrt( 9(x^2 - 16/9) ) dx and then: (1/3) * Integral of 1/ sqrt(x^2 - 16/9) dx?
As when i tried this, i got the incorrect answer.
Also, apologies if the format is difficult to understand, i tried to use latex, but got ' Latex Error: Compile failed'