Show that the derivative of the integral by looking at it geometrically and using the squeeze theorem: f(x) = ∫(^x, _0) (dt)/(1+t^2) is equal to arctan(x). Clarifying picture: http://i.imgur.com/OgprJ1Z.png I have done this algebraic but I'm struggling to do it with the conditions given. Sadly I don't find anything similar in my course book. Someone know how to do this? f(x)=∫(^x, _0) (dt)/(1+t^2) Let g(t) = 1/(1+t^2) Let h(x) = x h'(x) = 1 f'(x) = g ( h(x) ) h'(x) f'(x) = g(x) (1) f'(x) = 1/(1+x^2) The derivative of this integral is 1/(1+x^2) How about the geometrical part of it?