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Question Number 125911 by Lordose last updated on 15/Dec/20
∫0∞xln(1+x)1+x4dx
Answered by mathmax by abdo last updated on 15/Dec/20
A=∫0∞xln(1+x)1+x4dx=∫01xln(1+x)1+x4dx+∫1∞xln(x+1)1+x4dx∫1∞xln(x+1)1+x4dx=x=1t−∫01ln(1+1t)t(1+1t4)(−dtt2)=∫01ln(t+1)−ln(tt3+1tdt=∫01tln(t+1)−tlnt1+t4dt⇒A=2∫01xln(x+1)1+x4dx−∫01xlnx1+x4dxwehave∫01xlnx1+x4dx=∫01xln(x)∑n=0∞(−1)nx4ndx=∑n=0∞(−1)n∫01x4n+1ln(x)dx=∑n=0∞(−1)nunun=∫01x4n+1ln(x)dx=[14n+2x4n+2ln(x)]01−∫01x4n+14n+2dx=−1(4n+2)2⇒∫01xlnx1+x4dx=−∑n=0∞(−1)n4(2n+1)2=−k4(katalanconstant)∫01xln(x+1)1+x4dx=∫01xln(x+1)∑n=0∞(−1)nx4ndx=∑n=0∞(−1)n∫01x4n+1ln(x+1)dx=∑n=0∞(−1)nvnvn=∫01x4n+1ln(x+1)dx=x+1=t∫12(t−1)4n+1ln(t)dt=∫12ln(t)∑k=04n+1C4n+1ktk(−1)4n+1−kdt=−∑k=04n+1(−1)kC4n+1k∫12tkln(t)dtandbypsrts∫12tkln(t)dt=[1k+1tk+1ln(t)]12−∫12tk+1k+1dtt=2k+1ln(2)k+1−1k+1∫12tkdt=2k+1ln(2)k+1−1(k+1)2[tk+1]12=2k+1ln(2)k+1−1(k+1)2(2k+1−1)⇒vn=−∑k=04n+1(−1)kC4n+1k{2k+1ln(2)k+1−2k+1−1(k+1)2}....becontinued...
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