Showing that $\sum^{n}_{k=1}\frac{(-1)^{k}}{k} = -\log(2)+(-1)^n\cdot \int^{t=0}_{t=1}\frac{t^n}{(1+t)}$

Here's one way to do it $$\sum^{n}_{k=1}\frac{(-1)^k}{k}$$ $$=\sum^{n}_{k=1}(-1)^k\int_0^1\frac{t^k}{t+1}+\frac{t^{k-1}}{t+1}\ \mathrm dt$$ $$=\sum^{n}_{k=1}(-1)^k\int_0^1\frac{t^k+t^{k-1}}{t+1}\ \mathrm dt$$ $$=\sum^{n}_{k=1}(-1)^k\int_0^1t^{k-1}\ \mathrm dt$$ $$=\int_0^1\sum^{n}_{k=1}(-1)^kt^{k-1}\ \mathrm dt$$ $$=\int_0^1\frac{(-1)^nt^n-1}{t+1}\ \mathrm dt$$ $$=(-1)^n\int_0^1\frac{t^n}{t+1}\ \mathrm dt-\int_0^1\frac{1}{t+1}\ \mathrm dt$$ $$=(-1)^n\int_0^1\frac{t^n}{t+1}\ \mathrm dt-\ln 2$$

Let define $f(x)= \sum_{k=1}^{\infty}\frac {(-1)^k}kx^k$. This function is defined in $[0,1]$ and here is $C^\infty$.

Now $f'(x)= -\sum_{k=0}^{\infty} {(-1)^k}x^{k}= - \frac 1{1+x}$;

Integrating and valuating at $x=1$ with the condition $f(0)=0$ we found that $f(x)= -\log(1+x)$ so $f(1)=\sum_{k=1}^{\infty}\frac {(-1)^k}k=-\log(2)$.

If you break the series at $n$ we have the desired relation.