(FP3 question). Integrate 1/sqrt(3-4x-x^2).

There are two stages to this integral. Stage 1: Notice that there is a quadratic inside a square root in the denominator. We wish to look for a substitution which will reduce this problem to a known integral. To find this substitution we must mold the expression into a nicer form. Whenever you see a quadratic inside something nasty looking like this, a very common starting point is to complete the square giving us the integral dx/sqrt(7-(x+2)²). Stage 2: Now we wish to choose a trigonometric identity which can reduce this to a simpler problem. Notice that cos²(x)=1-sin²(x) may be used since this matches the form of the expression inside the square root. So if we let (x+2)²=7sin²(u) then dx/du=sqrt(7)cos(u) and our integral will become (by change of variables), sqrt(7)cos(u)du/sqrt(7-7sin²(u)) = sqrt(7)cos(u)du/(sqrt(7)cos(u)) = 1du. Now this easily integrates to u+C, where C is some constant. Rearrange (x+2)²=7sin²(u) to get u in terms of x and you will arrive at u=arcsin((x+2)/sqrt(7)). So our final answer is arcsin((x+2)/sqrt(7))+C, where C is some constant.