Another approach to the integral dx Vaz+ 1 is to use the circular trigonometric substitution x - tan(u). Then dx du sec^2(u) So I= sec2 (u) du = / sec(u)du. Vtan2 (u) + 1 clever trick. Multiply the function sec(u) by sec (u) + tan(u) This integral requires the use of a Since sec ( u ) +tan (u) (sec(u) + tan(u)) = then 1 = sec(u) + sec(u) tan(u) In(sec(u) + tan(u)) + c sec(u) + tan(u) In terms of x this becomes: I = Note: the Maple syntax for cos-(x) is cos (x) . The triangle below will help you to translate sec(u) into a function of r Vx2 + 1 sin(W) 72 41 cos(u)= tan(u) = x