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Integrate by parts

∫arctan(x) dx
Let
u = arctan(x) and dv = dx
du = dx/(x² + 1) and v = x

∫arctan(x) dx = x arctan(x) - ∫{x/(x² + 1)}dx
= x arctan(x) - (1/2)ln(x² + 1) + C
_______________________________

∫{1/[xln(x)]}dx
Let
u = ln(x)
du = dx/x

∫{1/[xln(x)]}dx = ∫(1/u)du = ln(u) + C = ln[ln(x)] + C

2007-02-09 14:43:23 · answer #1 · answered by Northstar 7 · 0 0

Arctan=tan^-1=cotso f(cot) is the integral. Sin(x)/x f(sinx(1/x), so just integrate .Remember 1/x is deriva.. of lnx

2007-02-09 00:36:27 · answer #2 · answered by Anonymous · 0 0

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