By using the substitution x = tan(u), find the integral of [1 / (x^2+1) dx] between the limits 1 and 0

First we need to change the limits, and by plugging in 1 and 0 to our substitution we find that the limits for u are arctan(1) and arctan(0) (or pi/4 and 0). Then we need to substitute for dx, and by differentiating our substitution equation we find that dx/du = sec^2(u), so dx = sec^2(u) du. Finally we substitute tan(u) for x in our integrand leaving us with [1 / (tan^2(u) +1)]. Therefore our new integral in u is [ sec^2(u) / (tan^2(u) + 1) du], between the limits pi/4 and 0. By recognising that tan^2(u) + 1 = sec^2(u), we find that actually the integrand entirely cancels, and leaves us with the integral of [1 du], which is of course [u]. Plugging in our limits of pi/4 and 0, we see that the final answer is pi/4. This is a hard question for A level but if a tutee of mine could answer this I'd feel confident they could answer any integration question in the exam.