How to integrate logarithm and power function?

I am trying to solve the following integral

$$\int_{0}^{1}(\ln(1+x))^2 x^{a-1}\,dx ; a>0.$$

I tried using partial functions but that didn't lead to anything.

Any suggestion?

The integral can be written as $$ I =\int_{0}^{1}\!dx\,\ln^2(1+x) x^{a-1}= \frac{\partial^2}{\partial b^2} \int_0^1 \!dx\,(1+x)^b x^{a-1} \Bigr|_{b=0} .$$

The remaining integral is an Euler type hypergeometric integral and can be solve by expanding $(1+x)^b$ in a Maclaurin series $$ (1+x)^b = \sum_{j=0}^\infty \binom{b}{j} x^j.$$ We obtain the result $$ I =\frac{\partial^2}{\partial b^2} \sum_{j=0}^\infty \binom{b}{j} \frac{1}{a+j} \Bigr|_{b=0}.$$

We have that $$ \binom{b}{j} = \frac{\Gamma(b+1)}{\Gamma(j+1) \Gamma(1+b-j)}$$ and $(d/dx)\ln \Gamma(x) =\psi(x)$ with $\psi(x)$ the digamma function. We obtain $$\frac{\partial}{\partial b} \ln\binom{b}{j} = \psi(b+1) - \psi(1+b -j) ; $$ and thus $$ I= \sum_{j=0}^\infty \binom{b}{j} \frac{ [\psi(b+1) - \psi(1+b -j)]^2 + \psi'(b+1) - \psi'(1+b -j) }{a+j} \Bigr|_{b=0}. \tag{1}$$

In a next step, we perform the limit $b\to 0$. For $j\in\mathbb{N}$, we have that $\binom{b}{j} = (-1)^{j+1}b/j +O(b^2)$. So we need to find, the contribution $\propto b^{-1}$ in the second factor of (1). We know that at close to integer values $j\in\mathbb{N}$, $$\psi(b-j) = -\frac{1}{b} + (H_{j} -\gamma) + O(b)$$ (this, can be derived from $\psi(z+1) = \psi(z) +1/z$) and $$\psi'(b-j) = \frac{1}{b^2} +O(1).$$ With that we obtain the limit ($j\in\mathbb{N}$) $$ \lim_{b\to0} \binom{b}{j}\biggl\{ [\psi(b+1) - \psi(1+b -j)]^2 + \psi'(b+1) - \psi'(1+b -j)\biggr\} = \frac{2 (-1)^j H_{j-1}}{j} $$ with the harmonic numbers $H_n =\sum_{k=1}^n k^{-1}$.

In conclusion, we obtain $$I = \sum_{j=1}^\infty \frac{2 (-1)^j H_{j-1}}{j(a+j)}.$$