不定积分 · 双元法初步 |
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该方法由 虚调子 提出. 双元法的本质是寻得一个或一组平方式. 其实只有两种情况: 要么 \(p^2+q^2=1\), 要么 \(p^2=q^2+1\). \(1.\quad\) \[\int \sqrt{1+x^2}\text dx \]利用 \[\int y\text dx=\frac12\int (y\text dx+x\text dy)+\frac12(y\text dx-x\text dy) \]就可以了. \(2.\quad\) \[\int\sqrt{\frac{x}{2+x^3}}\text dx \]构造双元 \[p=\sqrt{2+x^3},q=x^{\frac32} \]注意这种设法. \(3.\quad\) \[\int \frac{\ln(x+\sqrt{1+x^2})}{x^2}\text dx \]令 \(y=\sqrt{1+x^2}\) \[\begin{aligned} \int \frac{\ln(x+\sqrt{1+x^2})}{x^2}\text dx&=-\frac{\ln(x+y)}{x}+\int\frac1x\cdot\frac{\text d(x+y)}{x+y}\\ &=-\frac{\ln(x+y)}{x}+\int\frac{\text dy}{x^2}\\ &=-\frac{\ln(x+y)}{x}-\text{arctanh}(y) \end{aligned} \]这种属于直接换元. \(4.\quad\) \[\int \frac{\sqrt{1-x^2}}{\lambda^2+x^2-1}\text dx \]令 \(y=\sqrt{1-x^2}\) \[\int \frac{\sqrt{1-x^2}}{\lambda^2+x^2-1}\text dx=\int \frac{y\text dx}{\lambda^2-y^2} \]令 \(b=\dfrac{x}{y}\) 将其齐次化 \[\begin{aligned} \int \frac{y\text dx}{\lambda^2-y^2}=&\int \frac{y^4}{(\lambda^2(x^2+y^2)-y^2)(x^2+y^2)}\frac{\text dx}{y^3}\\ =&\int \frac{\text db}{(\lambda^2b^2+\lambda^2-1)(1+b^2)}\\ =&\int\frac{\lambda^2\text db}{\lambda^2b^2+\lambda^2-1}-\int\frac{\text db}{1+b^2} \end{aligned} \]这种通过双元第三公式齐次化的方法称为比值换元. \(5.\quad\) \[\int \frac{\text dx}{(x+1)\sqrt{x^2+1}} \]注意到恒等式 \[2(x^2+1)=(x+1)^2+(x-1)^2 \]不妨设 \[p=\frac{\sqrt2\sqrt{x^2+1}}{x+1},q=\frac{x-1}{x+1} \]则 \(p^2-q^2=1\) \[\begin{aligned} \int \frac{\text dx}{(x+1)\sqrt{x^2+1}}=&\frac{1}{\sqrt2}\int\frac{\text dq}{p}\\ =&\frac{1}{\sqrt2}\ln(p+q) \end{aligned} \]这种构造齐次恒等式换元的方法称为恒等式换元. 一般选取分母两项凑齐次. \(6.\quad\) \[\int\frac{x^2+1}{(x-1)^{\frac{5}{2}}\sqrt{1+x^3}}\text dx \]注意到恒等式 \[4(1+x^3)=3(x-1)^2(x+1)+(x+1)^3 \]不妨设 \[p=\frac{2}{x-1}\sqrt{\frac{x^3+1}{x+1}},q=\frac{x+1}{x-1},p^2=3+q^2 \]因此 \[\begin{aligned} \int\frac{x^2+1}{(x-1)^{\frac{5}{2}}\sqrt{1+x^3}}\text dx=&\int\frac{2\frac{x^2+1}{(x-1)^2}}{p\sqrt{q}}\frac{\text dx}{(x-1)^2}\\ =&-\int\frac{q^2+1}{p\sqrt{q}}\frac{\text dq}{2}\\ =&-\frac{1}{6}\int\frac{2q^2+p^2}{p\sqrt{q}}\text dq\\ =&-\frac{1}{3}\int\frac{2q\text dp+p\text dq}{2\sqrt{q}}\\ =&-\frac{1}{3}\int\text d(p\sqrt{q})\\ =&-\frac{1}{3}p\sqrt{q} \end{aligned} \]其中恒等式多由待定系数法给出. \(7.\quad\) \[\int \frac{\text dx}{x+\sqrt{x^2-x+1}} \]采用经典的 \(pqr\) 三换元： \[p=\sqrt{x-1}+\frac{1}{\sqrt{x-1}},q=\sqrt{x-1}-\frac{1}{\sqrt{x-1}},r=\sqrt{x+\frac{1}{x-1}} \]那么 \[p^2-q^2=4,p^2-r^2=1,r^2-q^2=3 \]\[\begin{aligned} \int \frac{\text dx}{x+\sqrt{x^2-x+1}}=&\int\frac{x-\sqrt{x^2-x+1}}{x-1}\text dx\\ =&x+\ln|x-1|-\int r\text d(p+q)\\ =&x+\ln|x-1|-r\sqrt{x-1}+\frac{1}{2}\ln(p+r)-\frac{3}{2}\ln(q+r) \end{aligned} \]这种被称为对勾换元. \(8.\quad\) \[\int \frac{\sqrt{1+x^4}}{1-x^4}\text dx \]设三元 \[p=\frac{1}{x}+x,q=\frac{1}{x}-x,r=\sqrt{x^2+\frac{1}{x^2}} \]那么 \[p^2-q^2=4,p^2-r^2=2,r^2-q^2=2 \]注意有平方差 \[pq=\frac{1}{x^2}-x^2 \]因此 \[\begin{aligned} \int \frac{\sqrt{1+x^4}}{1-x^4}\text dx=&\int \frac{r}{pq}\frac{\text dx}{x}\\ =&\int \frac{r}{pq}\frac{p+q}{2}\text d(\frac{p-q}{2})\\ =&\frac{1}{4}\int \frac{r}{pq}(q\text dp-p\text dq)\\ =&\frac{1}{4}\int\frac{r\text dp}{p}-\frac14\int \frac{r\text dq}{q}\\ =&\frac14(\int\frac{r^2\text dr}{r^2+2}-\int\frac{r^2\text dr}{r^2-2})\\ =&\frac{1}{2\sqrt2}(\text{arctanh}(\frac{r}{\sqrt2})-\arctan(\frac{r}{\sqrt2})) \end{aligned} \]\(9.\quad\) \[\int\frac{\text dx}{\sqrt{(x+x^{-1})^2-12}} \]仍然采用对勾三元 \[p=x+x^{-1},q=x-x^{-1},r=\sqrt{(x+x^{-1})^2-12} \]\[\begin{aligned} \int\frac{\text dx}{\sqrt{(x+x^{-1})^2-12}}=&\int\frac{\text d(p+q)}{2r}\\ =&\frac{1}{2}\ln((p+r)(q+r)) \end{aligned} \]\(10.\quad\) \[\int\frac{\sin^2x\cos x}{\sin x+\cos x}\text dx \]换一种设法： \[p=\cos x+\sin x,q=\cos x-\sin x,r=\sqrt{2\sin x\cos x} \]那么 \[\begin{aligned} \int\frac{\sin^2x\cos x}{\sin x+\cos x}\text dx=&\int\frac{\frac{r^2}{2}\cdot\frac{p-q}{2}}{p}\frac{\text dp}{q}\\ =&\frac14\int\frac{r^2}{q}\text dp-\frac14\int\frac{r^2}{p}\text dp\\ =&\frac{1}{4}\int\frac{\text dp}{q}-\frac{1}{4}\int q\text dp-\frac{1}{4}\int\frac{p^2-1}{p}\text dp\\ =&\frac14\ln p-\frac18(pq+p^2) \end{aligned} \]好啦休息一会儿，做一些小结~ 约定双元最简格式为 \[\int f \text dx \]定义双元常数为 \[y^2\pm x^2 \]有几串互推的等价式 \[\begin{aligned} &\frac{\text dx}{y^3}=\frac{1}{y^2\pm x^2}\text d(\frac{x}{y})\\ \Leftrightarrow&\frac{\text dx}{y}=\frac{y^2}{y^2\pm x^2}\text d(\frac{x}{y})=\frac{y\text dx-x\text dy}{y^2\pm x^2}\\ \Leftrightarrow&\int \frac{\text dx}{y}=\arctan(\frac{x}{y}) \text{(Re)}|| \ln(x+y) \text{(Im)} \end{aligned} \]以及双元点火公式 \[\begin{aligned} &\int y^3\text dx=\frac14 xy^3+\frac{1}{4}\cdot\frac{3}{2}(y^2\pm x^2)xy+\frac{1}{4}\cdot\frac{3}{2}(y^2\pm x^2)^2\int\frac{\text dx}{y}\\ &\int y^n\text dx=\frac{1}{n+1}xy^n+\frac{1}{n+1}\cdot\frac{n}{n-1}(y^2\pm x^2)xy^{n-2}+\cdots+\frac{1}{n+1}\cdot\frac{n}{n-1}\cdots\frac{3}{2}(y^2\pm x^2)^{\frac{n+1}{2}}\int\frac{\text dx}{y} \end{aligned} \]常见的几种处理情形: 直接换元 \(\quad\) \(y=\sqrt{1+x^2}\) 间接换元 \(\quad\) 比值换元 \(b=\dfrac{y}{x}\), 恒等式换元 \(2x^2+2=x^2+(x^2+2)\), 对勾换元 \(r=\sqrt{x+\dfrac{1}{x}+2}\) |
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