Taylor Series of ln(1+x) = x - x2/2 + x3/3 - ... (Derived from def'n of Maclaurin Series)
Taylor Series of ln(1+x2) = x2- x4/2 + x6/3 - ... (By substitution of x2 for x)
You can find the general integral of the TS by integrating each term
Integral of TS of ln(1+x2) = x3/3 - x5/10 + x7/21 - ... = Iapprox
Now just evaluate Iapprox from 0 to .5 i.e. Iapprox(.5) - Iapprox(0)
You can use the number of terms that get you close enough... (The numerical integral is .038867 for comparison)
Please consider a tutor. Take care
Doug C.
This is a graph prepared to help answer this question. but I will just append to Jacques D. response: desmos.com/calculator/sauhjftyjn09/06/22