I am stumped on an integration by parts problem. One of the integrals related to the Euler constant is:
$$\int_0^1 v^{n} log(1-v)dv = [ \frac{v^{n+1}-1}{n+1}log(1-v)]_0^1 - \int_0^1 \frac{v^{n+1}-1}{n+1}*\frac{1}{v-1}dv = -\frac{1}{n+1} \int_0^1 \frac{1-V^{n+1}}{1-V}dv$$ When I integrate this integral by parts I get the following instead :$$[{V^{n}}{n+1}log(1-v)]_0^1 - \frac{1}{n+1}\int_0^1\frac{V^{n}}{V-1}dv $$ I would appreciate anyone telling me where I am going wrong with this. Thanks in advance.
With $f(v)=\frac{1}{n+1} (v^{n+1}-1)$ and $g(v)=\log(1-v)$, we have $$(fg)'(v) = f'(v) g(v) + f(v) g'(v) = v^n \log(1-v) + \frac{1}{n+1} (v^{n+1}-1) \cdot \frac{1}{v-1}.$$ Integrate both sides and rearrange to get the first line in your post.
Your attempt is slightly wrong, but perhaps you meant $$\left[\frac{v^{n+1}}{n+1} \log(1-v)\right]_0^1 - \frac{1}{n+1} \int_0^1 \frac{v^{n+1}}{v-1} \mathop{dv}$$ Symbolically this is fine, but you get issues with the logarithm term as $v \to 1$.