find the integral from 0 to 0.5 for function [(arctan2x)/(1+4x^2)dx]?

The answer is as follows:

Substitute arctan(2x)=u and du=(2/(1+4x^2))dx
Now we have new integral
(1/2) Integral(u du)
But notice that limits must be changed from u=arctan(2*0)=0 to u=arctan(2*0.5)=0.7854
Now solve
(1/2) (0.7854^2 /2 - 0^2 /2) = 0.7854^2 /4 = 0.1542