I=∫(e2x+2ex−e−x−1)e(ex+e−x)dx
⇒I=∫(e2x+ex−1)e(ex+e−x)dx+∫(ex−e−x)eex+e−xdx
⇒I=∫(ex+1−e−x)eex+e−x+xdx+eex+e−x+c
ex+e−x+x=u
(ex−e−x+1)dx=du
I=eex+e−x+x+eex+e−x+c=eex+e−x(ex+1)+c
then g(x)=ex+1
g(0)=2
If ∫(e2x+2ex−e−x−1)e(ex+e−x)dx=g(x)e(ex+e−x)+c, where c is a constant of integration, then g(0) is
Held on 5 Sept 2020 · Verified 6 Jul 2026.
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