♠ t=tanx, x=atan(t), dx = dt/(1+t^2); I=∫y(t)*dt= ∫dt*ln(1+t)/(1+t^2) for 0<=t<=1; since the integral has no algebraic solution there are 2 ways: 1) to substitute with a tailor series; 2) to apply trapezoid method of integration; since tailor series for ln(1+t) has radius of convergence =1 the first way is rather problematic; thus nothing left but Simpson or trapez;

♣ thus I≈ {for k=1 until n-1}∑h*(y(t[k]) +(y(t[0]) +y(t[n])/2, where h=(t[n]-t[0])/n = 1/n is strip width, n is number of strips dependent on required accuracy of integration; assume accuracy 0.01;

♦ I=∑h*(y(t[k]) +0.25ln2;

let n=8; using MS Excel I=0.422;

let n=16; using MS Excel I=0.434;

let n=32; using MS Excel I=0.440;

let n=64; using MS Excel I=0.443;

let n=128; using MS Excel I=0.444