Is my evaluation of this integral correct? Here is my work.
$$I:=\int_{0}^{1}x^{n-1}\log(1+x)\,dx$$
$I.B.P$
$$I=\frac{1}{n}\log(2)-\frac{1}{n}\int_{0}^{1}\frac{x^n}{1+x}\,dx$$
Define:
$$J:=\int_{0}^{1}\frac{x^n}{1+x}\,dx$$
$$J=(-1)^n\int_{0}^{1}\frac{-1+(-x)^n+1}{1+x}\,dx$$
$$J=(-1)^n\log(2)+(-1)^{n-1}\int_{0}^{1}\frac{1-(-x)^n}{1+x}\,dx$$
$$J=(-1)^n\log(2)+(-1)^{n-1}\int_{0}^{1}\sum_{k=0}^{n}(-x)^k\,dx$$
$$J=(-1)^n\log(2)+(-1)^{n-1}\sum_{k=0}^{n}(-1)^k\int_{0}^{1}x^k\,dx$$
$$J=(-1)^n\log(2)+(-1)^{n-1}\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k}$$
$$I=\frac{1}{n}\log(2)-\frac{1}{n}[(-1)^n\log(2)+(-1)^{n-1}\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k}]$$
$$I=\frac{(1-(-1)^n)\log(2)}{n}+\frac{(-1)^n}{n}\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k}$$
And whether my work is correct or not, can this integral be expressed in terms of the harmonic numbers?
Your work seems correct so far, although I do not know how to finish it. Perhaps @Indianimperialist123's comment will help. Here is a method using series expansions.
Expanding the logarithm by its Taylor series around $x=1$, $$\begin{align*} \int_0^1x^{n-1}\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k}x^{k}\ dx&=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k}\int_0^1x^{k+n-1}\ dx \\ &=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k(n+k)} \end{align*}$$ splitting the alternating series into its even and odd parts, $$\begin{align*} \sum_{k=1}^\infty\frac{(-1)^{k+1}}{k(n+k)}=\sum_{k=1}^\infty\left(\frac{1}{(2k-1)(n+2k-1)}\right)-\sum_{k=1}^\infty\left(\frac{1}{2k(n+2k)}\right) \end{align*} $$ both series can be evaluated in terms of the Digamma function given the representation, $$\psi(x+1)=-\gamma+\sum_{k=1}^\infty\left(\frac{x}{k(x+k)}\right)$$ this yields a closed form, $$I_n=\frac{1}{2n}\left(\psi\left(\frac{n+1}{2}\right)-\psi\left(\frac{n}{2}+1\right)+\log(4)\right)$$ and the alternate form (using harmonic numbers) is obtained by the formula, $$\psi(x+1)=H_{x-1}-\gamma$$ so, $$I_n=\frac{1}{2n}\left(H_\frac{n-1}{2}-H_\frac{n}{2}+\log(4)\right).$$ Numerical check for $n=11$, $$\int_0^1x^{10}\log(1+x)\ dx=0.0590682135978\dots \\ \frac{H_{5}-H_{11/2}+\log(4)}{22}=0.0590682135978\dots$$