∫ex-1ex+1dx=f(x)+c,thenf(x)=
2logex+1
2loge2x–1
2logex+1–x
2loge2x+1
Explanation for the correct option:
Finding the given integral:
Given ∫ex-1ex+1dx=f(x)+c
Let us consider
u=exdu=exdx⇒duex=dx⇒duu=dx
Substituting in the given integral
=∫u−1u(u+1)du=∫2u−(u+1)u(u+1)dusubstitutingu=2u-u=2∫1u+1du−∫1udu=2lnu+1−lnu+C∫1u+1du=lnu+1=2lnex+1−x+CResubstitutingu=ex
Hence the correct answer is option (C).