Assuming the first term is [tex]\dfrac{e^y}x[/tex], we have by Green's theorem,
[tex]\displaystyle\int_C\left(\frac{e^y}x\,\mathrm dx+(e^y\ln x+2x)\,\mathrm dy\right)=\iint_R\left(\dfrac\partial{\partial x}\left(e^y\ln x+2x\right)-\dfrac\partial{\partial y}\left(\frac{e^y}x\right)\right)\,\mathrm dA[/tex]
[tex]=\displaystyle\int_{y=-1}^{y=1}\int_{x=y^4+1}^{x=2}\left(\frac{e^y}x+2-\frac{e^y}x\right)\,\mathrm dx\,\mathrm dy[/tex]
[tex]=\displaystyle2\int_{-1}^1\int_{y^4+1}^2\,\mathrm dx\,\mathrm dy[/tex]
[tex]=\displaystyle2\int_{-1}^1(1-y^4)\,\mathrm dy[/tex]
[tex]=\displaystyle4\int_0^1(1-y^4)\,\mathrm dy[/tex]
[tex]=\dfrac{16}5[/tex]