Answer: \(\dfrac{14}{3}-2\ln3\).

The two curves meet where

\(2+5\mathrm{e}^x=4-3\mathrm{e}^{2x}\).

Let \(u=\mathrm{e}^x\). Since \(\mathrm{e}^x\gt0\), we need \(u\gt0\). The equation becomes

\(2+5u=4-3u^2\).

Rearranging,

\(3u^2+5u-2=0\).

Factorising,

\((3u-1)(u+2)=0\).

Since \(u\gt0\), we take

\(u=\frac13\).

Thus

\(\mathrm{e}^x=\frac13\), so \(x=-\ln3\).

The other vertical boundary shown is the \(y\)-axis, \(x=0\). On the interval \(-\ln3\leqslant x\leqslant0\), the curve

\(y=2+5\mathrm{e}^x\)

lies above

\(y=4-3\mathrm{e}^{2x}\).

Therefore the shaded area is

\(\displaystyle \int_{-\ln3}^{0}\left[(2+5\mathrm{e}^x)-(4-3\mathrm{e}^{2x})\right]\,\mathrm{d}x\).

Simplifying the integrand,

\((2+5\mathrm{e}^x)-(4-3\mathrm{e}^{2x})=3\mathrm{e}^{2x}+5\mathrm{e}^x-2\).

So the area is

\(\displaystyle \int_{-\ln3}^{0}(3\mathrm{e}^{2x}+5\mathrm{e}^x-2)\,\mathrm{d}x\).

An antiderivative is

\(\dfrac32\mathrm{e}^{2x}+5\mathrm{e}^x-2x\).

Substituting the limits gives

\(\left[\dfrac32\mathrm{e}^{2x}+5\mathrm{e}^x-2x\right]_{-\ln3}^{0}\).

At \(x=0\), this is

\(\dfrac32+5=\dfrac{13}{2}\).

At \(x=-\ln3\), \(\mathrm{e}^x=\frac13\) and \(\mathrm{e}^{2x}=\frac19\), so the value is

\(\dfrac32\cdot\dfrac19+5\cdot\dfrac13-2(-\ln3)=\dfrac16+\dfrac53+2\ln3=\dfrac{11}{6}+2\ln3\).

Hence the area is

\(\dfrac{13}{2}-\left(\dfrac{11}{6}+2\ln3\right)=\dfrac{14}{3}-2\ln3\).