так?
[m]y=\frac{1}{5}x-\frac{1}{5}[/m]
[m]y=\frac{1}{5}[/m]
[m]dl=\sqrt{1+(y`(x))^2}dx=\sqrt{1+(\frac{1}{5})^2}dx=\frac{\sqrt{26}}{5}dx[/m]
[m] ∫_{ ∪AB}\frac{3}{(-2x+y-1)}dl= ∫_{0}^{5}\frac{3}{(-2x+\frac{1}{5}x-\frac{1}{5}-1)}\frac{\sqrt{26}}{5}dx= [/m]
это определенный интеграл:
[m]=3\sqrt{26}∫_{0}^{5}\frac{1}{(-10x+x-1-5)}dx=-3\sqrt{26}∫_{0}^{5}\frac{1}{(9x+6)}dx=-3\sqrt{26}∫_{0}^{5}\frac{3}{(9x+6)}dx=[/m]
[m]=-\frac{3\sqrt{26}}{9}(ln|9x+6|)|_{0}^{5}=-\frac{\sqrt{26}}{3}ln|45+6|+\frac{\sqrt{26}}{3}ln|0+6|=\frac{\sqrt{26}}{3}ln\frac{6}{51}=\frac{\sqrt{26}}{3}ln(\frac{2}{17}) [/m]