diff --git a/subjects/math/04_积分.md b/subjects/math/04_积分.md
index d862f3e..70c4c81 100644
--- a/subjects/math/04_积分.md
+++ b/subjects/math/04_积分.md
@@ -16,213 +16,193 @@ $$
 
 ---
 
-### 要点 02 - 分式型积分
+### 要点 02 - 分式型积分（$a, b > 0$）
 
 #### 基本公式
 
 $$
-\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \arctan \frac{x}{a} + C \quad (a > 0)
+\int \frac{dx}{a^2 x^2 + b^2} = \frac{1}{ab} \arctan \frac{ax}{b} + C
 $$
 
-**推导**（换元法）：令 $x = a \tan t$，则 $dx = a \sec^2 t \, dt$
+**推导**（换元法）：令 $t = \dfrac{a}{b} x$，则 $x = \dfrac{b}{a} t$，$dx = \dfrac{b}{a} dt$
 
 $$
-\int \frac{dx}{x^2 + a^2} = \int \frac{a \sec^2 t}{a^2 + a^2 \tan^2 t} \, dt
-$$
-
-$$
-= \int \frac{a \sec^2 t}{a^2 \sec^2 t} \, dt
-$$
-
-$$
-= \int \frac{dt}{a}
-$$
-
-$$
-= \frac{t}{a} + C
-$$
-
-$$
-= \frac{1}{a} \arctan \frac{x}{a} + C
+\int \frac{dx}{a^2 x^2 + b^2} = \int \frac{\frac{b}{a} dt}{b^2 t^2 + b^2}
+= \frac{1}{ab} \int \frac{dt}{t^2 + 1}
+= \frac{1}{ab} \arctan t + C
+= \frac{1}{ab} \arctan \frac{ax}{b} + C
 $$
 
 #### 推广形式
 
 $$
-\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \arctan \frac{x}{a} + C
+\int \frac{dx}{a^2 (x + c)^2 + b^2} = \frac{1}{ab} \arctan \frac{a(x + c)}{b} + C
 $$
 
 $$
-\int \frac{dx}{b^2 + (x + c)^2} = \frac{1}{b} \arctan \frac{x + c}{b} + C
+\int \frac{x \, dx}{a^2 x^2 + b^2} = \frac{1}{2a^2} \ln(a^2 x^2 + b^2) + C
 $$
 
 $$
-\int \frac{x \, dx}{x^2 + a^2} = \frac{1}{2} \ln(x^2 + a^2) + C
-$$
-
-$$
-\int \frac{dx}{(x^2 + a^2)^2} = \frac{x}{2a^2(x^2 + a^2)} + \frac{1}{2a^3} \arctan \frac{x}{a} + C
+\int \frac{dx}{(a^2 x^2 + b^2)^2} = \frac{x}{2b^2(a^2 x^2 + b^2)} + \frac{1}{2ab^3} \arctan \frac{ax}{b} + C
 $$
 
 ---
 
-### 要点 03 - 根号分式型积分
+### 要点 03 - 根号分式型积分（$a, b > 0$）
 
 #### 基本公式
 
-$$
-\int \frac{dx}{\sqrt{x^2 + a^2}} = \ln\left|x + \sqrt{x^2 + a^2}\right| + C
-$$
+**型 I**：$a^2 x^2 + b^2$
 
 $$
-\int \frac{dx}{\sqrt{x^2 + a^2}} = \operatorname{arsinh} \frac{x}{a} + C
+\int \frac{dx}{\sqrt{a^2 x^2 + b^2}} = \frac{1}{a} \ln\left|ax + \sqrt{a^2 x^2 + b^2}\right| + C
+= \frac{1}{a} \operatorname{arsinh} \frac{ax}{b} + C
 $$
 
-$$
-\int \frac{dx}{\sqrt{x^2 - a^2}} = \ln\left|x + \sqrt{x^2 - a^2}\right| + C \quad (|x| > |a|)
-$$
+**型 II**：$a^2 x^2 - b^2$
 
 $$
-\int \frac{dx}{\sqrt{x^2 - a^2}} = \operatorname{arcosh} \frac{x}{a} + C \quad (|x| > |a|)
+\int \frac{dx}{\sqrt{a^2 x^2 - b^2}} = \frac{1}{a} \ln\left|ax + \sqrt{a^2 x^2 - b^2}\right| + C \quad (|ax| > |b|)
+= \frac{1}{a} \operatorname{arcosh} \frac{ax}{b} + C
 $$
 
-$$
-\int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin \frac{x}{a} + C \quad (|x| < |a|)
-$$
+**型 III**：$b^2 - a^2 x^2$
 
 $$
-\int \frac{dx}{\sqrt{a^2 - x^2}} = -\arccos \frac{x}{a} + C \quad (|x| < |a|)
+\int \frac{dx}{\sqrt{b^2 - a^2 x^2}} = \frac{1}{a} \arcsin \frac{ax}{b} + C \quad (|ax| < |b|)
 $$
 
 #### 推导方法
 
-$$
-\begin{align}
-\int \frac{dx}{\sqrt{x^2 + a^2}}
-    &= \int \frac{a \cosh t}{a \cosh t} \, dt && (x = a \sinh t) \\
-    &= \int dt \\
-    &= t + C \\
-    &= \ln\left|x + \sqrt{x^2 + a^2}\right| + C
-\end{align}
-$$
+令 $t = ax$，则 $x = \dfrac{t}{a}$，$dx = \dfrac{dt}{a}$，化为标准形式后代入已知公式。
+
+**型 I**（$x = \frac{b}{a} \sinh t$ 或 $t = b \sinh u$）：
 
 $$
 \begin{align}
-\int \frac{dx}{\sqrt{x^2 - a^2}}
-    &= \int \frac{a \sinh t}{a \sinh t} \, dt && (x = a \cosh t) \\
-    &= \int dt \\
-    &= t + C \\
-    &= \ln\left|x + \sqrt{x^2 - a^2}\right| + C
+\int \frac{dx}{\sqrt{a^2 x^2 + b^2}}
+    &= \frac{1}{a} \int \frac{dt}{\sqrt{t^2 + b^2}}
+    = \frac{1}{a} \ln\left|t + \sqrt{t^2 + b^2}\right| + C \\
+    &= \frac{1}{a} \ln\left|ax + \sqrt{a^2 x^2 + b^2}\right| + C
 \end{align}
 $$
 
+**型 II**（$x = \frac{b}{a} \cosh t$）：
+
 $$
 \begin{align}
-\int \frac{dx}{\sqrt{a^2 - x^2}}
-    &= \int \frac{a \cos t}{a \cos t} \, dt && (x = a \sin t) \\
-    &= \int dt \\
-    &= t + C \\
-    &= \arcsin \frac{x}{a} + C
+\int \frac{dx}{\sqrt{a^2 x^2 - b^2}}
+    &= \frac{1}{a} \int \frac{dt}{\sqrt{t^2 - b^2}}
+    = \frac{1}{a} \ln\left|t + \sqrt{t^2 - b^2}\right| + C \quad (|t| > |b|) \\
+    &= \frac{1}{a} \ln\left|ax + \sqrt{a^2 x^2 - b^2}\right| + C
 \end{align}
 $$
 
-#### 等效形式
+**型 III**（$x = \frac{b}{a} \sin t$）：
 
 $$
-\int \frac{dx}{\sqrt{x^2 - a^2}} = \ln\left|\frac{x}{a} + \sqrt{\frac{x^2}{a^2} - 1}\right| + C
+\begin{align}
+\int \frac{dx}{\sqrt{b^2 - a^2 x^2}}
+    &= \frac{1}{a} \int \frac{dt}{\sqrt{b^2 - t^2}}
+    = \frac{1}{a} \arcsin \frac{t}{b} + C \quad (|t| < |b|) \\
+    &= \frac{1}{a} \arcsin \frac{ax}{b} + C
+\end{align}
 $$
 
 #### 推广形式
 
 $$
-\int \frac{dx}{\sqrt{(x + b)^2 + a^2}} = \ln\left|x + b + \sqrt{(x + b)^2 + a^2}\right| + C
+\int \frac{dx}{\sqrt{a^2 (x + c)^2 + b^2}} = \frac{1}{a} \ln\left|a(x + c) + \sqrt{a^2 (x + c)^2 + b^2}\right| + C
 $$
 
 $$
-\int \frac{dx}{\sqrt{(x + b)^2 - a^2}} = \ln\left|x + b + \sqrt{(x + b)^2 - a^2}\right| + C \quad (|x + b| > |a|)
+\int \frac{dx}{\sqrt{a^2 (x + c)^2 - b^2}} = \frac{1}{a} \ln\left|a(x + c) + \sqrt{a^2 (x + c)^2 - b^2}\right| + C \quad (|a(x + c)| > |b|)
 $$
 
 $$
-\int \frac{x \, dx}{\sqrt{x^2 + a^2}} = \sqrt{x^2 + a^2} + C
+\int \frac{x \, dx}{\sqrt{a^2 x^2 + b^2}} = \frac{1}{a^2} \sqrt{a^2 x^2 + b^2} + C
 $$
 
 $$
-\int \frac{x \, dx}{\sqrt{x^2 - a^2}} = \sqrt{x^2 - a^2} + C
-$$
-
-$$
-\int \sqrt{x^2 + a^2} \, dx = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\ln\left|x + \sqrt{x^2 + a^2}\right| + C
-$$
-
-$$
-\int \sqrt{x^2 - a^2} \, dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln\left|x + \sqrt{x^2 - a^2}\right| + C
+\int \frac{x \, dx}{\sqrt{a^2 x^2 - b^2}} = \frac{1}{a^2} \sqrt{a^2 x^2 - b^2} + C
 $$
 
 ---
 
-### 要点 04 - 根号二次型积分
+### 要点 04 - 根号二次型积分（$a, b > 0$）
 
 #### 基本公式
 
-$$
-\int \sqrt{x^2 + a^2} \, dx = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\ln\left|x + \sqrt{x^2 + a^2}\right| + C
-$$
+令 $t = ax$，统一化为标准形式后积分。
+
+**型 I**：$\sqrt{a^2 x^2 + b^2}$
 
 $$
-\int \sqrt{x^2 - a^2} \, dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln\left|x + \sqrt{x^2 - a^2}\right| + C \quad (|x| > |a|)
+\int \sqrt{a^2 x^2 + b^2} \, dx = \frac{x}{2}\sqrt{a^2 x^2 + b^2} + \frac{b^2}{2a} \ln\left|ax + \sqrt{a^2 x^2 + b^2}\right| + C
 $$
 
+**型 II**：$\sqrt{a^2 x^2 - b^2}$
+
 $$
-\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\arcsin\frac{x}{a} + C \quad (|x| < |a|)
+\int \sqrt{a^2 x^2 - b^2} \, dx = \frac{x}{2}\sqrt{a^2 x^2 - b^2} - \frac{b^2}{2a} \ln\left|ax + \sqrt{a^2 x^2 - b^2}\right| + C \quad (|ax| > |b|)
+$$
+
+**型 III**：$\sqrt{b^2 - a^2 x^2}$
+
+$$
+\int \sqrt{b^2 - a^2 x^2} \, dx = \frac{x}{2}\sqrt{b^2 - a^2 x^2} + \frac{b^2}{2a} \arcsin\frac{ax}{b} + C \quad (|ax| < |b|)
 $$
 
 #### 推导方法
 
-$$
-\begin{align}
-\int \sqrt{x^2 + a^2} \, dx
-    &= \int a \cosh t \cdot a \cosh t \, dt = a^2 \int \cosh^2 t \, dt && (x = a \sinh t) \\
-    &= a^2 \int \frac{\cosh 2t + 1}{2} \, dt = \frac{a^2}{2}\left(\frac{\sinh 2t}{2} + t\right) + C \\
-    &= \frac{a^2}{2}(\sinh t \cosh t + t) + C \\
-    &= \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\ln\left|x + \sqrt{x^2 + a^2}\right| + C
-\end{align}
-$$
+令 $t = ax$，则 $x = \frac{t}{a}$，$dx = \frac{dt}{a}$，化为对 $t$ 的标准形式。
+
+**型 I**（$t = b \sinh u$）：
 
 $$
 \begin{align}
-\int \sqrt{x^2 - a^2} \, dx
-    &= \int a \sinh t \cdot a \sinh t \, dt = a^2 \int \sinh^2 t \, dt && (x = a \cosh t) \\
-    &= a^2 \int \frac{\cosh 2t - 1}{2} \, dt = \frac{a^2}{2}\left(\frac{\sinh 2t}{2} - t\right) + C \\
-    &= \frac{a^2}{2}(\sinh t \cosh t - t) + C \\
-    &= \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln\left|x + \sqrt{x^2 - a^2}\right| + C
+\int \sqrt{a^2 x^2 + b^2} \, dx
+    &= \frac{1}{a} \int \sqrt{t^2 + b^2} \, dt
+    = \frac{1}{a} \left( \frac{t}{2}\sqrt{t^2 + b^2} + \frac{b^2}{2}\ln\left|t + \sqrt{t^2 + b^2}\right| \right) + C \\
+    &= \frac{x}{2}\sqrt{a^2 x^2 + b^2} + \frac{b^2}{2a} \ln\left|ax + \sqrt{a^2 x^2 + b^2}\right| + C
 \end{align}
 $$
 
+**型 II**（$t = b \cosh u$）：
+
 $$
 \begin{align}
-\int \sqrt{a^2 - x^2} \, dx
-    &= \int a \cos t \cdot a \cos t \, dt = a^2 \int \cos^2 t \, dt && (x = a \sin t) \\
-    &= \frac{a^2}{2}\left(t + \frac{\sin 2t}{2}\right) + C \\
-    &= \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\arcsin\frac{x}{a} + C
+\int \sqrt{a^2 x^2 - b^2} \, dx
+    &= \frac{1}{a} \int \sqrt{t^2 - b^2} \, dt
+    = \frac{1}{a} \left( \frac{t}{2}\sqrt{t^2 - b^2} - \frac{b^2}{2}\ln\left|t + \sqrt{t^2 - b^2}\right| \right) + C \\
+    &= \frac{x}{2}\sqrt{a^2 x^2 - b^2} - \frac{b^2}{2a} \ln\left|ax + \sqrt{a^2 x^2 - b^2}\right| + C
+\end{align}
+$$
+
+**型 III**（$t = b \sin u$）：
+
+$$
+\begin{align}
+\int \sqrt{b^2 - a^2 x^2} \, dx
+    &= \frac{1}{a} \int \sqrt{b^2 - t^2} \, dt
+    = \frac{1}{a} \left( \frac{t}{2}\sqrt{b^2 - t^2} + \frac{b^2}{2}\arcsin\frac{t}{b} \right) + C \\
+    &= \frac{x}{2}\sqrt{b^2 - a^2 x^2} + \frac{b^2}{2a} \arcsin\frac{ax}{b} + C
 \end{align}
 $$
 
 #### 推广形式
 
 $$
-\int (x + b)\sqrt{x^2 + a^2} \, dx = \frac{1}{3}(x + b)(x^2 + a^2)^{3/2} - \frac{b}{2}x\sqrt{x^2 + a^2} - \frac{ab^2}{2}\ln\left|x + \sqrt{x^2 + a^2}\right| + C
+\int x\sqrt{a^2 x^2 + b^2} \, dx = \frac{1}{3a^2}(a^2 x^2 + b^2)^{3/2} + C
 $$
 
 $$
-\int x\sqrt{x^2 + a^2} \, dx = \frac{1}{3}(x^2 + a^2)^{3/2} + C
+\int x\sqrt{a^2 x^2 - b^2} \, dx = \frac{1}{3a^2}(a^2 x^2 - b^2)^{3/2} + C
 $$
 
 $$
-\int x\sqrt{x^2 - a^2} \, dx = \frac{1}{3}(x^2 - a^2)^{3/2} + C
-$$
-
-$$
-\int x\sqrt{a^2 - x^2} \, dx = -\frac{1}{3}(a^2 - x^2)^{3/2} + C
+\int x\sqrt{b^2 - a^2 x^2} \, dx = -\frac{1}{3a^2}(b^2 - a^2 x^2)^{3/2} + C
 $$
 
 ---
@@ -863,9 +843,9 @@ $$
 - 定积分的定义
 - 黎曼和与积分的关系
 - 均匀分割技巧
-- $\frac{1}{x^2 + a^2}$ 型积分公式
-- $\frac{1}{\sqrt{x^2 \pm a^2}}$ 型积分公式
-- $\sqrt{x^2 \pm a^2}$ 型积分公式
+- $\frac{1}{a^2 x^2 + b^2}$ 型积分公式
+- $\frac{1}{\sqrt{a^2 x^2 \pm b^2}}$ 型积分公式
+- $\sqrt{a^2 x^2 \pm b^2}$ 型积分公式
 - 三角函数积分（降幂、万能代换、积化和差）
 - 第一类换元法（凑微分法）
 - 第二类换元法（变量代换法：三角/双曲/根式/倒代换）