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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...

A problem proposed by prof. Gadjev

Evaluate  \int_0^{\infty}\frac{\ln x}{x^2+2x+5}\, \text{d}x \textbf{Solution.} The indefinite integral is not expressible in elementary functions. Denote the integral by I. First make the change of variables \displaystyle x\to \frac{1}{x} to obtain  I=-\int_0^{\infty}\frac{\ln x}{5x^2+2x+1}\, \text{d}x\ \ In the original integral make the change of variables x\to 5x to obtain  I=\int_0^{\infty}\frac{\ln (5x)}{25x^2+10x+5}5\, \text{d}x=\int_0^{\infty}\frac{\ln x}{5x^2+2x+1}\, \text{d}x+\int_0^{\infty}\frac{\ln 5}{5x^2+2x+1}\, \text{d}x= -I+\int_0^{\infty}\frac{\ln 5}{5x^2+2x+1}\, \text{d}x whence  I=\frac{1}{2} \int_0^{\infty}\frac{\ln 5}{5x^2+2x+1}\, \text{d}x=\frac{1}{8}\arctan(2)\ln(5)