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编写小组/讲义/一元积分学（Part 1）.md

@@ -75,7 +75,7 @@ $\sqrt{x^2+a^2} = \sqrt{a^2\tan^2 t + a^2} = a\sec t$
 
 >[!note] 解析
 >令 $t=\sqrt[6]{x+1}$，则 $x=t^6-1$，$\mathrm dx=6t^5\mathrm dt$，则 
->$\begin{align}\int \frac{dx}{\sqrt[6]{x+1} - \sqrt[3]{x+1}} &= \int \frac{6t^5 dt}{t - t^2} \\&= 6\int \frac{t^4}{1-t} dt \\&= 6\int \left(-t^3 - t^2 - t - 1 + \frac{1}{1-t}\right) dt \\&= 6\left(-\frac{t^4}{4} - \frac{t^3}{3} - \frac{t^2}{2} - t - \ln|1-t|\right) + C \\&= -\frac{3}{2}\sqrt[3]{(x+1)^2} - 2\sqrt{x+1} - 3\sqrt[3]{x+1} - 6\sqrt[6]{x+1} - 6\ln|\sqrt[6]{x+1} -1| + C\end{align}$
+>$\begin{align}\int \frac{dx}{\sqrt[6]{x+1} - \sqrt[3]{x+1}} &= \int \frac{6t^5 dt}{t - t^2} \\&= 6\int \frac{t^4}{1-t} dt \\&= 6\int \left(-t^3 - t^2 - t - 1 + \frac{1}{1-t}\right) dt \\&= 6\left(-\frac{t^4}{4} - \frac{t^3}{3} - \frac{t^2}{2} - t - \ln|1-t|\right) + C \\&= -\frac{3}{2}\sqrt[3]{(x+1)^2} - 2\sqrt{x+1} - 3\sqrt[3]{x+1} - 6\sqrt[6]{x+1} -\\ 6\ln|\sqrt[6]{x+1} -1| + C\end{align}$
 
 >[!error] 待决策
 >##### 高次幂多项式与三角函数的转化