integrate integral of 1/(square root of 9x^2+6x-8)dx.
my work: integral dx/(suare root 9(x+ (1/3))^2 -7
u= x+(1/3) du=dx
integral of du/(9u^2-7) u=square root 7/3 sec theta squareroot (7sectheta^2 - 7) =square root 7theta^2.
need help? i dont think i dont thing i did any of this right.