Another approach to the integral I = VI +1 is to use the circular trigonometric substitution r = tan(u). Then dx du sec^2(u) So I = sec- (u ) du - sec(u) du. tan? (u) + 1 This integral requires the use of a trick. Multiply the function sec(u) by sec(u) +tan(u) clever Since sec(u) +tan(u) du (sec(u) + tan(u)) = then I = sec(u) + sec(u) tan(u) -In(sec(u) + tan(u)) + C. sec(u) + tan(u) In terms of x this becomes: BE +C. Note: the Maple syntax for cos (x) is cos (x) . The triangle below will help you to translate sec(u) into a function of x. sin(u) /x2 + 1 /2441 1 cos( u ) = tan(u) = x