If $\int\left(\mathrm{e}^{2 \mathrm{x}}+2 \mathrm{e}^{\mathrm{x}}-\mathrm{e}^{-\mathrm{x}}-1\right) \mathrm{e}^{\left(\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}\right)} \mathrm{dx}=\mathrm{g}(\mathrm{x}) \mathrm{e}^{\left(\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}\right)}+\mathrm{c}$ where c is a constant of integration, then $\mathrm{g}(0)$ is equal to:

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Indefinite Integrals

\int (e^{2 x} + 2 e^{x} - e^{- x} - 1) e^{(e^{x} + e^{- x})} dx = g (x) e^{(e^{x} + e^{- x})} + c

where c is a constant of integration, then

g (0)

is equal to:

Answer: (B)

Split the integrand so that part of it is an exact derivative. Regrouping, the integral is

\int [(e^{2 x} + e^{x} - 1) + (e^{x} - e^{- x})] e^{(e^{x} + e^{- x})} d x

. The piece

(e^{x} - e^{- x}) e^{(e^{x} + e^{- x})}

integrates directly to

e^{(e^{x} + e^{- x})}

, and combining it with the rest the antiderivative is

(e^{x} + 1) e^{(e^{x} + e^{- x})} + c

. Thus

g (x) = e^{x} + 1

and

g (0) = 2

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Single correct option