Video narration
This integral looks familiar, but the fraction inside the square root does not match a standard formula. Start by rationalising it: multiply inside the radical by (x+2)/(x+2). This changes the fraction to (x+2)2/(x2−4), so the radical becomes (x+2)/√(x2−4), on the relevant interval. Here is the key turn: split the numerator. Now, (x+2)/√(x2−4) = x/√(x2−4) + 2/√(x2−4). Therefore, I is the sum of ∫x/√(x2−4) dx and 2∫dx/√(x2−4). The reason this split helps is that the x in the first numerator matches the derivative of x2−4, while the second is exactly the logarithmic form you already know. For the first integral, use substitution t=√(x2−4). Then x dx=t dt, so it becomes ∫dt=t, giving √(x2−4). For the second, use the standard formula for ∫dx/√(x2−a2): it is ln|x+√(x2−a2)|. With a=2, multiply by 2. The absolute value is required. Recombining, and adding the constant of integration C, gives: I=√(x2−4)+2 ln|x+√(x2−4)|+C.