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.