Answer
$$\int_0^1\frac{e^z+1}{e^z+z}dz=\ln(e+1)$$
Work Step by Step
To evaluate the integral
$$\int_0^1\frac{e^z+1}{e^z+z}dz$$
we will use substitution $e^z+z=t$ which gives us $(e^z+1)dz=dt$. The integration bounds would be: for $z=0$ we have $t=1$ and for $z=1$ we have $t=e+1$. Putting this into the integral we get:
$$\int_0^1\frac{e^z+1}{e^z+z}dz=\int_1^{e+1}\frac{1}{t}dx=\left.\ln|t|\right|_1^{e+1}=\ln(e+1)-\ln1=\ln(e+1)$$