Somemathhere

 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)&#

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  $$I=\frac{1}{2} \int_0^{\infty}\frac{\ln 5}{5x^2+2x+1}\, \text{d}x=\frac{1}{8}\arctan(2)\ln(5)$$