微积分学/定积分/反常积分

例題

$\begin{array}{cc}& 1.\underset{0}{\overset{3}{\int }}\frac{1}{x^2-6x+5}dx\\ & =\underset{0}{\overset{3}{\int }}\frac{1}{(x-1)(x-5)}dx\\ & =\underset{0}{\overset{3}{\int }}\frac{A}{(x-1)}+\frac{B}{(x-5)}dx\\ & A=\frac{-1}{4}B=\frac{1}{4}\\ & =\underset{0}{\overset{3}{\int }}\frac{-1}{4}\frac{1}{(x-1)}+\frac{1}{4}\frac{(x-5)}{d}x\\ & =\frac{-1}{4}\underset{0}{\overset{3}{\int }}\frac{1}{(x-1)}dx+\frac{1}{4}\underset{0}{\overset{3}{\int }}\frac{1}{(x-5)}dx\\ & =\frac{-1}{4}(\underset{b\to 1^{-}}{\lim }\underset{0}{\overset{b}{\int }}\frac{1}{(x-1)}dx+\underset{b\to 1^{+}}{\lim }\underset{b}{\overset{3}{\int }}\frac{1}{(x-1)}dx)+\frac{1}{4}[\ln (x-5){]}_0^3\\ & =\frac{-1}{4}\left(\underset{b\to 1^{-}}{\lim }\right[\ln (x-1){]}_0^b+\underset{b\to 1^{+}}{\lim }[\ln (x-1){]}_b^3)+\frac{1}{4}\ln \frac{-2}{-5}\\ & =\frac{-1}{4}(\underset{b\to 1^{-}}{\lim }\ln \frac{b-1}{-1}+\underset{b\to 1^{+}}{\lim }\ln \frac{2}{b-1})+\frac{1}{4}\ln \frac{2}{5}\\ & =\frac{-1}{4}(\ln 0^{+}+\ln 1+\ln 2-\ln 0^{+})+\frac{1}{4}\ln \frac{2}{5}\\ & =\frac{-1}{4}(\mathrm{\infty }-\mathrm{\infty }+\ln 2)+\frac{1}{4}\ln \frac{2}{5}\\ & =\text{undefined}\\ & \\ & 2.\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{1}{1+x^2}dx\\ & \tan \frac{\pi }{2}=\mathrm{\infty }\tan \frac{-\pi }{2}=-\mathrm{\infty }\\ & =[{\tan }^{-1}x{]}_{-\mathrm{\infty }}^{\mathrm{\infty }}=\frac{1}{\pi }+\frac{1}{\pi }=\pi \end{array}$