What does $\pi$ have to do with $\ln$ integrals ?

Before we evaluate the integral, let us first notice the following propositions which would help us to evaluate the given integral.

Proposition 1 : $\sum_{n=1}^\infty H_n x^n=-\frac{\ln(1-x)}{1-x}\quad;\quad\text{for}\ |x|<1,$ where $H_n$ is the $n$ -th harmonic number .

Proof :

Since $H_n-H_{n-1}=\frac{1}{n}$ for $n\in\mathbb{Z}_+$ , then multiplying both sides by $x^n$ yields $\begin{aligned} H_n x^n-H_{n-1}x^n&=\frac{x^n}{n}\\ \sum_{n=1}^\infty H_n x^n-\sum_{n=1}^\infty H_{n-1}x^n&=\sum_{n=1}^\infty\frac{x^n}{n}\\ \sum_{n=1}^\infty H_n x^n-x\sum_{n=1}^\infty H_{n-1}x^{n-1}&=-\ln(1-x)\\ (1-x)\sum_{n=1}^\infty H_n x^n&=-\ln(1-x)\qquad\text{Q.E.D.} \end{aligned}$ where $H_0=0$ , and $H_1=1$ .

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Proposition 2 : $\sum_{n=1}^\infty\frac{H_n}{n^3}=\frac{\pi^4}{72}.$ Proof :

We have $H_n=\sum_{m=1}^\infty\left(\frac{1}{m}-\frac{1}{m+n}\right)= \sum_{m=1}^\infty \frac{n}{m(m+n)}.$ Therefore $\sum_{n=1}^\infty\frac{H_n}{n^3}=\sum_{m,n\geq1}\frac{1}{n^2m(m+n)} =\sum_{m,n\geq1}\frac{1}{m^2n(n+m)}.$ Taking the half sum we obtain $\begin{aligned} \sum_{n=1}^\infty\frac{H_n}{n^3}&=\frac{1}{2}\sum_{m,n\geq1}\frac{1}{mn(m+n)}\left(\frac{1}{m}+\frac{1}{n}\right)\\ &=\frac{1}{2}\sum_{m,n\geq1}\frac{1}{m^2n^2}\\ &=\frac{1}{2}\zeta^2(2)\\ &=\frac{1}{2}\left(\frac{\pi^2}{6}\right)^2\qquad\qquad\qquad\text{Q.E.D.} \end{aligned}$

Proposition 3 : $\int_0^1 x^\alpha \ln^n x\ dx=\frac{(-1)^n n!}{(\alpha+1)^{n+1}}, \qquad\text{for }\ n=0,1,2,\ldots$

Proof :

Note that $\int_0^1 x^\alpha\ dx=\frac1{\alpha+1},\qquad\text{for }\ \alpha>-1.$ Differentiating $n$ times yields $\int_0^1 \frac{\partial^n}{\partial\alpha^n}\left(x^\alpha\right)\ dx=\int_0^1 x^\alpha \ln^n x\ dx=\frac{(-1)^n n!}{(\alpha+1)^{n+1}}, \qquad\text{for }\ n=0,1,2,\ldots\qquad\text{Q.E.D.}$

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Now, we will evaluate the given integral. Using the property $\int_a^b f(x)\ dx=\int_a^b f(a+b-x)\ dx$ and also the previous propositions, then the given integral turns out to be $\begin{aligned} \int_0^1\frac{\ln x\ln^2(1-x)}{x}\ dx&=\int_0^1\frac{\ln(1-x)\ln^2 x}{1-x}\ dx\\ &=-\int_0^1\sum_{n=1}^\infty H_n x^n\ln^2 x\ dx\\ &=-\sum_{n=1}^\infty H_n\int_0^1 x^n\ln^2 x\ dx\\ &=-2\sum_{n=1}^\infty\frac{H_n}{(n+1)^3}\\ &=-2\sum_{n=1}^\infty\frac{1}{(n+1)^3}\left(H_{n+1}-\frac{1}{n+1}\right)\\ &=2\left[\sum_{n=1}^\infty\frac{1}{(n+1)^4}-\sum_{n=1}^\infty\frac{H_{n+1}}{(n+1)^3}\right]\\ &=2\left[\sum_{n=1}^\infty\frac{1}{n^4}-\sum_{n=1}^\infty\frac{H_{n}}{n^3}\right]\\ &=2\left[\zeta(4)-\frac{\pi^4}{72}\right]\\ &=2\left[\frac{\pi^4}{90}-\frac{\pi^4}{72}\right]\\ &=\large\color{#3D99F6}{-\frac{\pi^4}{180}}. \end{aligned}$ Thus, $\large a+b+c=1+4+180=\boxed{\color{#3D99F6}{185}}$ .