∫(1+(logx)2)2(logx−1)2dx=∫[(1+(logx)2)1−(1+(logx)2)22logx]dx=∫[1+t2et−(1+t2)22tet]dt put logx=t⇒dx=etdt∫et[1+t21−(1+t2)22t]dt=1+t2et+c=1+(logx)2x+c
JEE Main 2005 — Mathematics Calculus
∫{(1+(logx)2(logx−1)}2dx is equal to
Held on 30 Apr 2005 · Verified 6 Jul 2026.
(logx)2+1logx+C
x2+1x+C
1+x2xex+C
(logx)2+1x+C
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