∫e4logex+5e3logex−7e2logexe3loge2x+5e2loge2xdx,x>0
=∫x4+5x3−7x2(2x)3+5(2x)2dx=∫x2(x2+5x−7)4x2(2x+5)dx
=4∫(x2+5x−7)d(x2+5x−7)=4loge∣x2+5x−7∣+c
JEE Main 2021 — Mathematics Calculus
The integral ∫e4logex+5e3logex−7e2logexe3loge2x+5e2loge2xdx,x>0, is equal to
(where c is a constant of integration)
Held on 25 Feb 2021 · Verified 6 Jul 2026.
loge∣x2+5x−7∣+c
4loge∣x2+5x−7∣+c
41loge∣x2+5x−7∣+c
logex2+5x−7+c
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