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Question Number 144402 by rexford last updated on 25/Jun/21
Commented by justtry last updated on 25/Jun/21
Commented by rexford last updated on 27/Jun/21
thanksverymuchforyourtimeIamverygrateful
Answered by mathmax by abdo last updated on 25/Jun/21
Ψ=∫dxx227x2+6x−2⇒Δ′=32−27.(−2)=9+54=63⇒x1=−3+6327x2=−3−6327⇒Ψ=∫dxx227(x−x1)(x−x2)=127∫dxx2x−x1x−x2=x−x1=t127∫2tdt(t2+x1)2tt2+x1−x2=227∫dt(t2+x1)2t2+x1−x2wehavex1−x2=26327sowedothecha7gementt=26327shy⇒Ψ=227∫λ0chydy(λ02sh2y+x1)2λ0chy=227∫dy(λ02sh2y+x1)2(λ0=26327)....becontinued...
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