yes.
Let u = arctan(x)
du = dx/(x^2 + 1)
dv = x dx
v = x^2/2
That gets rid of the pesky trig stuff, and you're just left with a rational function, which brings back the arctan after integrating!
How do I integrate x(arctanx) dx? Is it some sort of integration by parts? Can you please show steps? Thanks in advance!
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