1/(1+x^5) = 1 - x^5 + x^10 - R(x)

with remainder R(x)>0, less than 0.2^15 on (0, 0.2)

Therefore:

I = INT 0_0.2 1/(1 + x^5) dx = integral_{x=0 to 0.2} 1 - x^5 + x^10 - R(x) dx

..................................= 0.2 - (0.2)^6/6 + (0.2)^11/11 + epsilon

where |epsilon| < (0.2)^16.

So I ~= 0.194444 to six decimal places

RonL