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**choovuck** for any fixed z choose a branch of logarithm so that [z,z+1] is not on the cut (e.g. take the cut $\displaystyle (-\infty,0]$) if z is not negative, and take the cut $\displaystyle [0,\infty)$ if it is (i.e if $\displaystyle z<1$)). for this branch $\displaystyle \log(z+t)$ for $\displaystyle t\in[0,1]$ is a primitive for $\displaystyle \frac{1}{z+t}$, so $\displaystyle \int_0^1 \frac{1}{z+t}dt=\log(z+1)-\log z$

writing the answer in the form $\displaystyle ln((z+1)/z)$, as you did, is not correct. you should write as $\displaystyle \log(z+1)-\log z$ where you explain what branch of logarithm you're taking for each particular z