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Question Number 65320 by hovea cw last updated on 28/Jul/19
Commented by mathmax by abdo last updated on 28/Jul/19
letI=∫011+xlnxdxchangement1+x=tgive1+x=t2⇒x=t2−1andI=∫12tln(t2−1)(2t)dt=2∫12t2ln(t2−1)dtbypartsu′=t2andv=ln(t2−1)⇒∫12t2ln(t2−1)dt=[(t33−13)ln(t2−1)]12−∫12(t3−13)2tt2−1dt=(223−13)×0−limt→1t3−13ln(t2−1)−23∫12t4−tt2−1dt=0−0−23∫12t4−tt2−1dt∫12t4−tt2−1dt=∫12t(t3−1)t2−1dt=∫12t(t−1)(t2+t+1)(t−1)(t+1)dt=∫12t3+t2+tt+1dt=∫12t2(t+1)−t2+t2+tt+1dt=∫12t2dt+∫12tdtt+1=[t33]12+∫12(1−1t+1)dt=223−13+2−1−[ln∣t+1∣]12=532−43−ln(1+2)+ln(2)⇒I=−43{532−43+ln(21+2)}I=−2092+169−43ln(21+2)
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