If (e^2x+2e^x-e^-x-1)e^(e^x+e^-x),dx=g(x)e^(e^x+e^-x)+c, where c is a constant of integrat…

Question

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

Accepted Answer

$$I=\int(e^{2x}+2e^x-e^{-x}-1)e^{(e^x+e^{-x})}dx=\int(e^x+1-e^{-x})e^x\cdot e^{(e^x+e^{-x})}dx+\int(e^x-e^{-x})e^{(e^x+e^{-x})}dx$$

Regrouping: $=\displaystyle\int(e^x-e^{-x}+1)e^{(e^x+e^{-x}+x)}dx+e^{(e^x+e^{-x})}$

Let $e^x+e^{-x}+x=t\Rightarrow(e^x-e^{-x}+1)dx=dt$:
$$I=\int e^tdt+e^{(e^x+e^{-x})}=e^t+e^{(e^x+e^{-x})}+C=e^{(e^x+e^{-x}+x)}+e^{(e^x+e^{-x})}+C=(e^x+1)e^{(e^x+e^{-x})}+C$$

So $g(x)=1+e^x$, and $g(0)=2$

Solutions · 1