<s>[INST] Integral $\int_0^1 \frac{\sqrt x \ln x} {x^2 - x+1}dx$ I am trying to evaluate $$I=\int_0^1 \frac{\sqrt x \ln x} {x^2 - x+1}dx=\int_0^1 \frac{\sqrt x (1+x)\ln x} {1+x^3}dx$$ Now if we expand into  geometric series: $$I=\sum_{n=0}^{\infty} (-1)^n \int_0^1 (x^{3/2}+x^{1 /2})x^{3n}\ln x dx$$ Also since $$I(k) =\int_0^1 x^kdx=\frac{1} {k+1}$$ Giving: $$I'(k) =\int_0^1 x^k\ln x dx=-\frac{1} {(k+1)^2 }$$ so using this we get $$I=\sum_{n=0}^{\infty} (-1)^{n+1}\left(\frac{1} {(6n+3)^2 }+\frac{1} {(6n+1) ^2 }\right)$$ Now when I plug this into wolfram-alpha the result differs from the value of the integral, also if I multiply by a  half it is really close to it. Where did I go wrong? 
Edit: Looks like I forgot a 2 in the denominator and to add $+1$ from $I'(k) $ and the correct series should be:$$I=\frac{4}{36}\sum_{n=0}^{\infty} (-1)^{n+1}\left(\frac{1} {(n+5/6)^2 }+\frac{1} {(n+1/2) ^2 }\right)$$ The second one is just $-4G$ where $G$ is the Catalan constant and can you show me how to transform the sum into a closed form? Trigamma or hurwitz zeta function as wolfram alpha gives as a solution. Many thanks in advance! 
 [/INST] $$ \frac{\sqrt{x}}{x^2-x+1} = \sum_{k=0}^\infty (-1)^k \left(x^{3k+1/2} + x^{3k + 3/2}\right) $$ so
$$ \frac{\sqrt{x} \ln(x)}{x^2-x+1} = \left.\sum_{k=0}^\infty  (-1)^k \dfrac{d}{dp}\left(x^{3k+1/2+p} + x^{3k+3/2+p}\right)\right|_{p=0}$$
$$ \eqalign{\int_0^1 \frac{\sqrt{x} \ln(x)}{x^2-x+1} \; dx &= \sum_{k=0}^\infty (-1)^k \left.\dfrac{d}{dp} \left( \frac{1}{3k+3/2+p} + \frac{1}{3k+5/2+p} \right)\right|_{p=0}\cr &= \sum_{k=0}^\infty (-1)^{k+1} \left( \frac{1}{(3k+3/2)^2} + \frac{1}{(3k+5/2)^2}\right)\cr}$$
EDIT: In terms of $\zeta(2,v) = \sum_{j=0}^\infty 1/(j+v)^2$, we can write this as
$$ \frac{1}{36} \left(-\zeta(2,1/4) -\zeta(2,5/12) + \zeta(2,3/4)+\zeta(2,11/12)\right) $$
since for $k=2j$, $3k+3/2 = 6 (j+1/4)$, for $k=2j+1$, $3k+3/2 = 6 (j+3/4)$, etc.
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