Question

Найти значение определенного интеграла

Answer 1

$\int _1^2\, \frac{dx}{x^4}=\frac{x^{-3}}{-3}\, |_1^2=-\frac{1}{3}(\frac{1}{8}-1)=\frac{7}{24}\\\\\\\int _5^{5\sqrt3}\frac{dx}{25+x^2}=\frac{1}{5}arctg\frac{x}{5}\, |_5^{5\sqrt3}=\frac{1}{5}(arctg\sqrt3-arctg1)=\frac{1}{5}(\frac{\pi}{3}-\frac{\pi}{4})=\frac{\pi}{60}$