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a
loglogx+2logx+3+c
b
loglogx−log3logx−log2+c
c
loglogx−3logx−2+c
d
loglogx+3logx+2+c
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Correct option is C
Put logx=u Then dxx=du Hence Given =∫dxx(logx−2)(logx−3)=∫duu−2u−3 Using partial fractions, 1(u−2)(u−3)=Au−2+Bu−31=Au−3+Bu−2 Put u=2⇒−1=A Put u=3⇒1=B=∫du(u−2)(u−3)=∫1u−3−1u−2du=log(u−3)−log(u−2)=logu−3u−2+c=loglogx−3logx−2+cSimilar Questions
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