I'm going to assume that you meant to use parenthesis around portions of your derivative function and that it is supposed to be:
That means to find the function f(x) that we must integrate:
Let's first break this fraction (and the associated integral) into two:
f(x) = ∫1/(x2+1)dx - ∫(2x)/(x2+1)dx
To make it easier to explain, let's call the problem f(x) = A - B where A = ∫1/(x2+1)dx and B = ∫(2x)/(x2+1)dx
Integrating A is easy if you remember that the derivative of arctan(x) = 1/(x2 + 1) which means that the integral of 1/(x2 + 1)dx is arctan(x).
To Integrate B, let u = x2 + 1, the du/dx = 2x so du = 2xdx meaning that we can write the integral as:
∫1/u du and the integral of this is ln(u). Back-substituting, we get ln(x2 + 1).
So f(x) = A - B becomes:
f(x) = arctan(x) - ln(x2 + 1) + C
To find the value of C, we can plug in the point (0, 6) so:
6 = arctan(0) - ln(02 + 1) + C
6 = 0 - 0 + C
C = 6
f(x) = arctan(x) - ln(x2 + 1) + 6