微积分(A)随缘一题[21]

计算：\(\int \sqrt{x^2-1}dx\)

考虑分部积分法：

\[\begin{aligned} &\int \sqrt{x^2-1}dx = x\sqrt{x^2-1}-\int \frac{(x^2-1)+1}{\sqrt{x^2-1}}dx \\ &\int \sqrt{x^2-1}dx=\frac{x\sqrt{x^2-1}-\int \frac{dx}{\sqrt{x^2-1}}}{2}=\frac{x\sqrt{x^2-1}-\ln|x+\sqrt{x^2-1}|}{2}+C \end{aligned} \]

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考虑双元法：

\[\begin{aligned} &\int \sqrt{x^2-1}dx \xlongequal[y^2=x^2-1]{y=\sqrt{x^2-1}}\int ydx =I=\frac{x\sqrt{x^2-1}-\ln |x+\sqrt{x^2-1}|}{2}+C \\ &I=\int ydx,J=\int xdy \\ &I+J=\int xdy+ydx =\int d(xy)=xy+C \\ &I-J=\int ydx-xdy=\int ydx-\frac{x^2}{y}dx=-\int \frac{dx}{y}=-\int \frac{d(x+y)}{x+y}=-\ln |x+y|+C \\ \end{aligned} \]

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\[\int \frac{dx}{\sqrt{x^2\pm a^2}} \xlongequal[p^2=x^2+a^2,pdp=xdx]{p=\sqrt{x^2+a^2}}\int \frac{dx}{p}=\int \frac{d(x+p)}{p+x}=\ln |p+x|+C=\ln |\sqrt{x^2 \pm a^2}+x|+C \]

\[\begin{aligned} \int \frac{dx}{\sqrt{a^2-x^2}} \xlongequal[p^2+x^2=a^2,pdp+xdx=0]{p=\sqrt{a^2-x^2}} &\int \frac{dx}{p} \\ =& \int \frac{pdx-xdp}{p^2+x^2}=\int \frac{p^2}{p^2+x^2}d\frac{x}{p} \\ =&\arctan \frac{x}{p}+C=\arctan \frac{x}{\sqrt{a^2-x^2}}+C=\arcsin \frac{x}{a}+C \end{aligned} \]

\[\int \frac{dx}{x^2-a^2}=\frac{1}{2a}\int \left(\frac{1}{x-a}-\frac{1}{x+a}\right)dx=\frac{1}{2a}\ln \left| \frac{x-a}{x+a} \right| + C \]

\[\int \frac{dx}{x^2+a^2} = \frac{1}{a}\int \frac{d \frac{x}{a}}{1+\frac{x^2}{a^2}}=\frac{\arctan \frac{x}{a}}{a}+C \]

\[\int \frac{dx}{x^2+a^2} \xlongequal{x=a\tan t}\int \frac{a\sec^2t dt}{a^2\sec^2t}=\frac{t}{a}+C=\frac{\arctan \frac{x}{a}}{a}+C \]