integrate (2x)/(x^2+1) dx with limits 1, 0

Firstly we notice that the numerator is the derivative of the denominator so we can use integration by subsitution method. Setting u=x^(2)+1. We can differentiate this to get du/dx=2x Subbing in dx=du/2x . This cancels out the 2x in the function we are trying to integrate. We are left with the integral of 1/u du. However we must not forget to change the limits as they as with respect to x and not u. We can sub in the limits into u=x^2+1. When x=1, u=(1)^2+1 = 2When x = 0, u=(0)^2+1 = 1Therefor our new limits are 2,1 If we integrate 1/u du with limits 2,1 we get ln(u) + c (this is a standard rule) Subbing in the limits we get: ln(2) - ln(1) = 0.693147...We have worked out the area under the curve f(x)=(2x)/(x^2+1) between x=1 and x=0