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Question Number 104028 by mathmax by abdo last updated on 19/Jul/20
calculate∫20+∞dx(x−18)19(x−19)18
Answered by Dwaipayan Shikari last updated on 19/Jul/20
∫20+∞(x−19)dx(x2−37x+342)1912∫20+∞2x−37(x2−37x+342)19dx−1(x2−37+342)19−136[1(x2−37x+342)18]20+∞−12∫1(x−18)19(x−19)19136.1218−{12∫20+∞1(x−18)19(x−19)19}→Ia136.218−Iacontinue......
Answered by mathmax by abdo last updated on 19/Jul/20
A=∫20+∞dx(x−18)19(x−19)18changementx−19=tgiveA=∫1+∞dt(t+1)19t18=∫1∞dt(tt+1)18(t+1)37wedothechangementtt+1=u⇒t=ut+u⇒(1−u)t=u⇒t=u1−u⇒dtdu=1−u+u(1−u)2=1(1−u)2t+1=u1−u+1=u+1−u1−u=11−u⇒A=∫121du(1−u)2u18(11−u)37=∫121(1−u)37(1−u)2u18du=∫121(1−u)35u18du=−∫121(u−1)35u18du=−∫121∑k=035C35kuk(−1)35−ku18du=∑k=035(−1)kC35k∫121uk−18du=∑k=0andk≠1735(−1)kC35k[1k−17uk−17]121−C3517[ln∣u∣]121A=∑k=0andk≠1735(−1)kC35kk−17{1−12k−17}−ln(2)C3517
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