Answer
$\int_{0}^{1}\frac{e^z+1}{e^z+z}~dz = ln~(e+1)$
Work Step by Step
$\int_{0}^{1}\frac{e^z+1}{e^z+z}~dz$
Let $u = e^z+z$
$\frac{du}{dz} = e^z+1$
$dz = \frac{du}{e^z+1}$
When $x = 0$, then $u = 1$
When $x = 1$, then $u = e+1$
$\int_{1}^{e+1} \frac{e^z+1}{u}\cdot \frac{du}{e^z+1}$
$=\int_{1}^{e+1} \frac{du}{u}$
$=ln~u~\vert_{1}^{e+1}$
$=ln~(e+1)-ln~(1)$
$=ln~(e+1)$
