A problem told by Zhivko Petrov

 The following integral was proposed for homework to Applied Math, by Zhivko Petrov. 

\textbf{Problem}. Evaluate

\int_{0}^1\frac{\ln(1-x+x^2)}{x^2-x}\text{d} x.

Later I communicated the problem to prof. Gadjev and he proposed a neat solution (to be presented later).

Now we elaborate a solution, on an idea proposed by prof. Babev. I would like to thank the afformentioned people, as well as David Petrov for pointing me to that idea.

\textbf{Solution}. Introduce I(y)=\int_0^1\frac{\ln(1-y(x-x^2))}{x^2-x}\text{d}x.

We need to find I(1). Clearly I(0)=0. Using differentiation under the integral sign we obtain 

I'(y)=\int_0^1\frac{1}{1-y(x-x^2)}\text{d}x.

Assuming that y\in [0,1], one can easily integrate the last to obtain 

I'(y)=\frac{4 \arcsin\left(\frac{\sqrt{y}}{2}\right)}{\sqrt{y(4-y) }}.

The  latter is very easy to integrate (for example making the change y=4t^2) in order to obtain

I'(y)=\left(4 \arcsin\left(\frac{\sqrt{y}}{2}\right)^2\right)'.

Thus 

I(1)=\int_0^1I'(y)\text{d}y+I(0)= 4 \arcsin\left(\frac{\sqrt{1}}{2}\right)^2-4 \arcsin\left(\frac{\sqrt{0}}{2}\right)^2+0=4 \arcsin\left(\frac{1}{2}\right)^2=\frac{\pi^2}{9}.

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