[m] ∫_{1} ^{ +∞ }\frac{16x}{16x^4-1}dx=lim _{A → + ∞ }2\cdot ∫_{1} ^{ A }\frac{8x}{16x^4-1}dx=2\cdot lim _{A → + ∞ }∫_{1} ^{ A }\frac{d(4x^2)}{(4x^2)^2-1}dx=2\frac{1}{2}lim_{A →+ ∞ } ln|\frac{4x^2-1}{4x^2+1}|)|_{1} ^{ A }=lim_{A →+ ∞ } ln|\frac{4x^2-1}{4x^2+1}|-ln|\frac{4\cdot 1^2-1}{4\cdot 1^2+1}|=ln1-ln\frac{3}{5}[/m]
Сходится
[m] ∫_{1} ^{3 }\frac{dx}{\sqrt{x^2-6x+9}}=lim _{ ε → +0 }∫_{1} ^{3- ε }\frac{dx}{\sqrt{(x-3)^2}}=lim _{ ε → +0 }∫_{1} ^{3- ε }\frac{dx}{|x-3|}=lim _{ ε → +0 }∫_{1} ^{3- ε }\frac{dx}{(-(x-3))}=-lim _{ ε → +0 }ln|x-3|)|_{1} ^{3- ε }=[/m]