Integral from 0 to 1 of: [ln(x+1)dx] / [(x^2)+1] ?

Answer 1

I'm afraid you will have to resort to numerical methods
to compute this, because ln(x+1)/(x²+1) does not
have an elementary antiderivative. Its integral
involves the dilogarithm function.
Let's make it a bit easier to work with.
Try integration by parts:
Let
u = ln(x+1) dv = 1/(x²+1) dx
du = 1/(x+1) v = arctan x

Then ∫ ln(x+1) dx/(x²+1) =
ln(x+1) arctan x - ∫ arctan(x) dx/(x+1),
(all with limits 0 to 1)
So we get

π/4 ln 2 - ∫ arctan x dx/(x+1)
I suggest evaluating this last integral by expanding
arctan x in a Taylor series, dividing the result by
x+1 and calculating it to sufficiently many decimal
places.