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Question Number 31099 by abdo imad last updated on 02/Mar/18
find∫0∞arctan(2x)−arctanxxdx.
Commented by abdo imad last updated on 04/Mar/18
I=limξ→+∞I(ξ)/I(ξ)=∫0ξartan(2x)−arctanxxdxI(ξ)=∫0ξarctan(2x)xdx−∫0ξarctanxxdxch.2x=tgive∫0ξarctan(2x)xdx=∫02ξarctan(t)t2dt2=∫02ξarctanttdt⇒I(ξ)=∫02ξarctanxxdx−∫0ξarctanxxdx=∫ξ2ξarctanxxdxbut∃c∈]ξ,2ξ[/I(ξ)=artanξ∫ξ2ξdtt=ln(2)arctanξ⇒limξ→+∞I(ξ)=π2ln2⇒I=π2ln(2).
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