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素材/整合素材/高数素材/不定积分大集合——技巧篇.md

@@ -24,11 +24,18 @@ $$\int \frac{1}{(x^2+1)\sqrt{x^2-1}} dx$$
 解：令 $x = \sec t$，则 $dx = \sec t \tan t dt$
 
 $$\begin{align*}
-\int \frac{1}{(x^2+1)\sqrt{x^2-1}} dx &= \int \frac{\sec t \tan t}{(\sec^2 t+1)\tan t} dt \\
-&= \int \frac{\cos t}{\cos^2 t+1} dt \\
+\int \frac{1}{(x^2+1)\sqrt{x^2-1}} dx 
+= \int \frac{\sec t \tan t}{(\sec^2 t+1)\tan t} dt 
+= \int \frac{\cos t}{\cos^2 t+1} dt \\
+\end{align*}$$
+令 $u = sin t$
+  
+由 $x = \sec t$,得 $\cos t = \frac{1}{x}$，所以
+
+$$\begin{align*}\sin t = \sqrt{1 - \frac{1}{x^2}} = \frac{\sqrt{x^2-1}}{x} \quad \\(\text{当 }x>1\text{ 时取正})\\
 
-&=  \frac{1}{2\sqrt{2}} \ln\left| \frac{\sqrt{x^2-1} + x\sqrt{2}}{\sqrt{x^2-1} - x\sqrt{2}} \right| + C\\
-\end{align*}\\
+原式等于  \frac{1}{2\sqrt{2}} \ln\left| \frac{\sqrt{x^2-1} + x\sqrt{2}}{\sqrt{x^2-1} - x\sqrt{2}} \right| + C\\
+\end{align*}$$
 $$
 【2.2】