Integral

[man111]

Evaluation of [tex]\displaystyle \int^{1}_{0}\frac{\ln(x+1)}{x^2+5x+6}dx[/tex]

[vezni]

Означаваме $I=\int_0^1\frac{\ln(x+1)}{x^2+5x+6}\,dx$. След смяната $x=\frac{1-t}{1+t}$ получаваме
$I=\int_0^1\frac{\ln\frac{2}{1+t}}{t^2+5t+6}\,dt=\ln 2\int_0^1\frac{dt}{t^2+5t+6}-\int_0^1\frac{\ln(1+t)}{t^2+5t+6}\,dt=\ln 2\int_0^1\frac{dt}{t^2+5t+6}-I$
$\Rightarrow I=\frac{\ln 2}{2}\int_0^1\frac{dt}{t^2+5t+6}$.
$\int_0^1\frac{dt}{t^2+5t+6}=\int_0^1\left(\frac{1}{t+2}-\frac{1}{t+3}\right)\,dt=\ln\frac{t+2}{t+3}\Bigg\rvert_0^1=\ln\frac 98$
$\Rightarrow I=\frac{\ln 2\ln\frac 98}{2}$