If I=∫ e-x log (ex+1) d x, then I equals

Question

If I=∫ e-x log (ex+1) d x, then I equals (A) x+(e-x+1) log (ex+1)+C (B) x+(ex+1) log (ex+1)+C (C) x-(e-x+1) log (ex+1)+C (D) None of these

Tardigrade Admin · Accepted Answer

I =-e-x log (ex+1)+∫ (e-x ex/ex+1) d x =-e-x log (ex+1)+∫ (e-x/e-x+1) d x =-e-x log (ex+1)- log (e-x+1)+C =-e-x log (ex+1)- log (1+ex)+x+ C =-(e-x+1) log (ex+1)+x+ C

Q. If $I = \int e^{- x} lo g (e^{x} + 1) d x$ , then $I$ equals

$x + (e^{- x} + 1) lo g (e^{x} + 1) + C$

$x + (e^{x} + 1) lo g (e^{x} + 1) + C$

$x - (e^{- x} + 1) lo g (e^{x} + 1) + C$

None of these

Q. If I=∫e−xlog(ex+1)dx, then I equals

x+(e−x+1)log(ex+1)+C

x+(ex+1)log(ex+1)+C

x−(e−x+1)log(ex+1)+C

None of these

I=−e−xlog(ex+1)+∫ex+1e−xex​dx =−e−xlog(ex+1)+∫e−x+1e−x​dx =−e−xlog(ex+1)−log(e−x+1)+C =−e−xlog(ex+1)−log(1+ex)+x+C =−(e−x+1)log(ex+1)+x+C

Q. If $I = \int e^{- x} lo g (e^{x} + 1) d x$ , then $I$ equals

$x + (e^{- x} + 1) lo g (e^{x} + 1) + C$

$x + (e^{x} + 1) lo g (e^{x} + 1) + C$

$x - (e^{- x} + 1) lo g (e^{x} + 1) + C$