1. ∫ 0 3 1 x 2 − 6 x + 5 d x = ∫ 0 3 1 ( x − 1 ) ( x − 5 ) d x = ∫ 0 3 A ( x − 1 ) + B ( x − 5 ) d x A = − 1 4 B = 1 4 = ∫ 0 3 − 1 4 1 ( x − 1 ) + 1 4 ( x − 5 ) d x = − 1 4 ∫ 0 3 1 ( x − 1 ) d x + 1 4 ∫ 0 3 1 ( x − 5 ) d x = − 1 4 ( lim b → 1 − ∫ 0 b 1 ( x − 1 ) d x + lim b → 1 + ∫ b 3 1 ( x − 1 ) d x ) + 1 4 [ ln ( x − 5 ) ] 0 3 = − 1 4 ( lim b → 1 − [ ln ( x − 1 ) ] 0 b + lim b → 1 + [ ln ( x − 1 ) ] b 3 ) + 1 4 ln − 2 − 5 = − 1 4 ( lim b → 1 − ln b − 1 − 1 + lim b → 1 + ln 2 b − 1 ) + 1 4 ln 2 5 = − 1 4 ( ln 0 + + ln 1 + ln 2 − ln 0 + ) + 1 4 ln 2 5 = − 1 4 ( ∞ − ∞ + ln 2 ) + 1 4 ln 2 5 = undefined 2. ∫ − ∞ ∞ 1 1 + x 2 d x tan π 2 = ∞ tan − π 2 = − ∞ = [ tan − 1 x ] − ∞ ∞ = 1 π + 1 π = π {\displaystyle {\begin{aligned}&1.\;\int _{0}^{3}{\frac {1}{x^{2}-6x+5}}dx\\&=\int _{0}^{3}{\frac {1}{(x-1)(x-5)}}dx\\&=\int _{0}^{3}{\frac {A}{(x-1)}}+{\frac {B}{(x-5)}}dx\\&\qquad A={\frac {-1}{4}}\qquad B={\frac {1}{4}}\\&=\int _{0}^{3}{\frac {-1}{4}}{\frac {1}{(x-1)}}+{\frac {1}{4}}{\frac {(x-5)}{d}}x\\&={\frac {-1}{4}}\int _{0}^{3}{\frac {1}{(x-1)}}dx+{\frac {1}{4}}\int _{0}^{3}{\frac {1}{(x-5)}}dx\\&={\frac {-1}{4}}\left(\lim _{b\to 1^{-}}\int _{0}^{b}{\frac {1}{(x-1)}}dx+\lim _{b\to 1^{+}}\int _{b}^{3}{\frac {1}{(x-1)}}dx\right)+{\frac {1}{4}}{\bigg [}\ln(x-5){\bigg ]}_{0}^{3}\\&={\frac {-1}{4}}\left(\lim _{b\to 1^{-}}{\bigg [}\ln(x-1){\bigg ]}_{0}^{b}+\lim _{b\to 1^{+}}{\bigg [}\ln(x-1){\bigg ]}_{b}^{3}\right)+{\frac {1}{4}}\ln {\frac {-2}{-5}}\\&={\frac {-1}{4}}\left(\lim _{b\to 1^{-}}\ln {\frac {b-1}{-1}}+\lim _{b\to 1^{+}}\ln {\frac {2}{b-1}}\right)+{\frac {1}{4}}\ln {\frac {2}{5}}\\&={\frac {-1}{4}}\left(\ln 0^{+}+\ln 1+\ln 2-\ln 0^{+}\right)+{\frac {1}{4}}\ln {\frac {2}{5}}\\&={\frac {-1}{4}}\left(\infty -\infty +\ln 2\right)+{\frac {1}{4}}\ln {\frac {2}{5}}\\&={\text{undefined}}\\&\\&2.\;\int _{-\infty }^{\infty }{\frac {1}{1+x^{2}}}dx\\&\quad \tan {\frac {\pi }{2}}=\infty \qquad \tan {\frac {-\pi }{2}}=-\infty \\&={\Big [}\tan ^{-1}x{\Big ]}_{-\infty }^{\infty }={\frac {1}{\pi }}+{\frac {1}{\pi }}=\pi \end{aligned}}}
![{\displaystyle {\begin{aligned}&1.\;\int _{0}^{3}{\frac {1}{x^{2}-6x+5}}dx\\&=\int _{0}^{3}{\frac {1}{(x-1)(x-5)}}dx\\&=\int _{0}^{3}{\frac {A}{(x-1)}}+{\frac {B}{(x-5)}}dx\\&\qquad A={\frac {-1}{4}}\qquad B={\frac {1}{4}}\\&=\int _{0}^{3}{\frac {-1}{4}}{\frac {1}{(x-1)}}+{\frac {1}{4}}{\frac {(x-5)}{d}}x\\&={\frac {-1}{4}}\int _{0}^{3}{\frac {1}{(x-1)}}dx+{\frac {1}{4}}\int _{0}^{3}{\frac {1}{(x-5)}}dx\\&={\frac {-1}{4}}\left(\lim _{b\to 1^{-}}\int _{0}^{b}{\frac {1}{(x-1)}}dx+\lim _{b\to 1^{+}}\int _{b}^{3}{\frac {1}{(x-1)}}dx\right)+{\frac {1}{4}}{\bigg [}\ln(x-5){\bigg ]}_{0}^{3}\\&={\frac {-1}{4}}\left(\lim _{b\to 1^{-}}{\bigg [}\ln(x-1){\bigg ]}_{0}^{b}+\lim _{b\to 1^{+}}{\bigg [}\ln(x-1){\bigg ]}_{b}^{3}\right)+{\frac {1}{4}}\ln {\frac {-2}{-5}}\\&={\frac {-1}{4}}\left(\lim _{b\to 1^{-}}\ln {\frac {b-1}{-1}}+\lim _{b\to 1^{+}}\ln {\frac {2}{b-1}}\right)+{\frac {1}{4}}\ln {\frac {2}{5}}\\&={\frac {-1}{4}}\left(\ln 0^{+}+\ln 1+\ln 2-\ln 0^{+}\right)+{\frac {1}{4}}\ln {\frac {2}{5}}\\&={\frac {-1}{4}}\left(\infty -\infty +\ln 2\right)+{\frac {1}{4}}\ln {\frac {2}{5}}\\&={\text{undefined}}\\&\\&2.\;\int _{-\infty }^{\infty }{\frac {1}{1+x^{2}}}dx\\&\quad \tan {\frac {\pi }{2}}=\infty \qquad \tan {\frac {-\pi }{2}}=-\infty \\&={\Big [}\tan ^{-1}x{\Big ]}_{-\infty }^{\infty }={\frac {1}{\pi }}+{\frac {1}{\pi }}=\pi \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9722a17f00d50c369569453ab9d8ad463df46d5e)