# Integrating ln(x+1)/(x^2+1) using recursive integration by parts

Hi, I need to find ∫ln(x+1)/(x^2+1)dx

I think it might involve recursive integration by parts, so first I set:

u=ln(x+1) dv = 1/(x^2+1)dx
du=1/(x+1)dx v=ArcTan(x)

∫ln(x+1)/(x^2+1)dx = ArcTan(x)Ln(x+1) - ∫ArcTan(x)/(x+1)dx

Then I integrated by parts again, so

u=1/(x+1) dv=ArcTan(x)
du=-1/(x+1)^2dx v=x*ArcTan(x)-Ln(x^2+1)/(x+1)

∫Ln(x+1)/(x^2+1)dx = ArcTan(x)Ln(x+1) -1/(x+1)(x*ArcTan(x)-Ln(x^2+1)/2)+∫x*ArcTan(x)/(x+1)^2-Ln(x^2+1)/(2(x+1)^2)

my problem here is that neither my ArcTan term nor my natural log term that I'm attempting to integrate can be rewritten in terms of either of the two previous integrals I've done. I have the natural log of (x^2+1) instead of (x+1) and my ArcTan term is divided by (x+1)^2 rather than (x+1). So I guess I could continue integration by parts, but that's just getting real ugly real fast. Any ideas?