From 6b07a9701078cb2f8775fe1bf4271346582f46c3 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E5=88=98=E6=9F=AF=E5=A6=A4?= <2503393720@qq.com>
Date: Wed, 28 Jan 2026 16:30:19 +0800
Subject: [PATCH] vault backup: 2026-01-28 16:30:19

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+1. 指数函数相关积分
+【1.1】
+
+$$\int \frac{e^x - e^{-x}}{e^x + e^{-x}} dx$$
+
+解：
+
+$$\int \frac{e^x - e^{-x}}{e^x + e^{-x}} dx = \int \frac{d(e^x + e^{-x})}{e^x + e^{-x}} = \ln|e^x + e^{-x}| + C$$
+【1.2】
+
+$$\int \frac{e^{2x} - e^x}{e^{2x} + 1} dx$$
+
+解：
+
+$$\begin{align*}
+\int \frac{e^{2x} - e^x}{e^{2x} + 1} dx &= \int \frac{e^{2x}}{e^{2x} + 1} dx - \int \frac{e^x}{e^{2x} + 1} dx \\
+&= \frac{1}{2}\ln(e^{2x} + 1) - \arctan e^x + C
+\end{align*}$$
+2. 根式与倒数相关积分
+【2.1】
+
+$$\int \frac{1}{(x^2+1)\sqrt{x^2-1}} dx$$
+
+解：令 $x = \sec t$，则 $dx = \sec t \tan t dt$
+
+$$\begin{align*}
+\int \frac{1}{(x^2+1)\sqrt{x^2-1}} dx &= \int \frac{\sec t \tan t}{(\sec^2 t+1)\tan t} dt \\
+&= \int \frac{\cos t}{\cos^2 t+1} dt \\
+&= \arctan(\sin t) + C \\
+&= \arctan\left(\frac{\sqrt{x^2-1}}{x}\right) + C
+\end{align*}$$
+【2.2】
+
+$$\int \frac{1}{x\sqrt{x^2-1}} dx$$
+
+解：令 $x = \sec t$，则 $dx = \sec t \tan t dt$
+
+$$\int \frac{1}{x\sqrt{x^2-1}} dx = \int \frac{\sec t \tan t}{\sec t \tan t} dt = \int 1 dt = t + C = \arccos\frac{1}{x} + C$$
+【2.3】
+
+$$\int \frac{1}{\sqrt{x(1-x)}} dx$$
+
+解：
+
+$$\int \frac{1}{\sqrt{x(1-x)}} dx = 2\int \frac{1}{\sqrt{1-(\sqrt{x})^2}} d\sqrt{x} = 2\arcsin\sqrt{x} + C$$
+【2.4】
+
+$$\int \frac{1}{\sqrt{1+e^{2x}}} dx$$
+
+解：令 $t = \sqrt{1+e^{2x}}$，则 $e^{2x} = t^2-1$，$dx = \frac{t}{t^2-1} dt$
+
+
+$$\begin{align*}
+\int \frac{1}{\sqrt{1+e^{2x}}} dx &= \int \frac{1}{t} \cdot \frac{t}{t^2-1} dt \\
+&= \frac{1}{2}\ln\left|\frac{t-1}{t+1}\right| + C \\
+&= \ln\left|\sqrt{1+e^{2x}} - 1\right| - x + C
+\end{align*}$$
+【2.5】
+
+$$\int \frac{x^3}{(x^4+1)^2} dx$$
+
+解：
+
+$$\int \frac{x^3}{(x^4+1)^2} dx = \frac{1}{4}\int \frac{d(x^4+1)}{(x^4+1)^2} = -\frac{1}{4(x^4+1)} + C$$
+3. 三角函数积分
+【3.1】
+
+$$\int \frac{1}{1+\cos x} dx$$
+
+解：
+
+$$\int \frac{1}{1+\cos x} dx = \int \frac{1-\cos x}{\sin^2 x} dx = \int \csc^2 x dx - \int \csc x \cot x dx = -\cot x + \csc x + C$$
+【3.2】
+
+$$\int \frac{1}{1+\sin x} dx
+$$
+解：
+
+$$\int \frac{1}{1+\sin x} dx = \int \frac{1-\sin x}{\cos^2 x} dx = \int \sec^2 x dx - \int \sec x \tan x dx = \tan x - \sec x + C$$
+4. 分式积分
+【4.1】
+
+$$\int \frac{1}{x^2-1} dx$$
+
+解：
+
+$$\int \frac{1}{x^2-1} dx = \frac{1}{2}\int \left(\frac{1}{x-1} - \frac{1}{x+1}\right) dx = \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C$$
+【4.2】
+
+$$\int \frac{1}{x^2+1} dx$$
+
+解：
+
+$$\int \frac{1}{x^2+1} dx = \arctan x + C$$
+5. 三角乘积与幂次积分
+【5.1】
+
+$$\int \frac{1}{\cos x (\sin x + \cos x)} dx$$
+
+解：
+
+$$\begin{align*}
+\int \frac{1}{\cos x (\sin x + \cos x)} dx &= \int \frac{\sec^2 x}{\tan x + 1} dx \\
+&= \int \frac{d(\tan x + 1)}{\tan x + 1} \\
+&= \ln|\tan x + 1| + C
+\end{align*}$$
+【5.2】
+
+$$\int \frac{\sin x}{\sin x - \cos x} dx$$
+
+解：
+
+$$\begin{align*}
+\int \frac{\sin x}{\sin x - \cos x} dx &= \frac{1}{2}\int \frac{(\sin x - \cos x) + (\sin x + \cos x)}{\sin x - \cos x} dx \\
+&= \frac{1}{2}\int 1 dx + \frac{1}{2}\int \frac{\sin x + \cos x}{\sin x - \cos x} dx \\
+&= \frac{1}{2}x + \frac{1}{2}\ln|\sin x - \cos x| + C
+\end{align*}$$
+【5.3】
+
+$$\int \sec x \tan x dx$$
+
+解：
+
+$$\int \sec x \tan x dx = \sec x + C$$
+6. 高次三角函数积分
+【6.1】
+
+$$\int \frac{1}{(\sin x - 1)\cos x} dx$$
+
+解：
+
+$$\begin{align*}
+\int \frac{1}{(\sin x - 1)\cos x} dx &= \int \frac{\sin x + 1}{(\sin x - 1)(\sin x + 1)\cos x} dx \\
+&= \int \frac{\sin x + 1}{-\cos^3 x} dx \\
+&= -\int \sec^2 x \tan x dx - \int \sec^3 x dx \\
+&= -\frac{1}{2}\sec^2 x - \frac{1}{2}(\sec x \tan x + \ln|\sec x + \tan x|) + C
+\end{align*}$$
+7. 三角分式积分
+【7.1】
+
+$$\int \frac{1}{1+\sin x + \cos x} dx$$
+
+解：令 $t = \tan\frac{x}{2}$，则 $\sin x = \frac{2t}{1+t^2}$，$\cos x = \frac{1-t^2}{1+t^2}$，$dx = \frac{2}{1+t^2} dt$
+
+$$\begin{align*}
+\int \frac{1}{1+\sin x + \cos x} dx &= \int \frac{1}{1+\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}} \cdot \frac{2}{1+t^2} dt \\
+&= \int \frac{1}{1+t} dt \\
+&= \ln|1+t| + C \\
+&= \ln\left|1+\tan\frac{x}{2}\right| + C
+\end{align*}$$
+【7.2】
+
+$$\int \frac{1}{\sin x \cos x} dx$$
+
+解：
+
+$$\int \frac{1}{\sin x \cos x} dx = \int \csc 2x \cdot 2 dx = \ln|\tan x| + C$$
+8. 三角幂次积分
+【8.1】
+
+$$\int \frac{1}{\sin^2 x \cos^2 x} dx
+$$
+解：
+
+$$\int \frac{1}{\sin^2 x \cos^2 x} dx = \int \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} dx = \int \sec^2 x dx + \int \csc^2 x dx = \tan x - \cot x + C$$
+【8.2】
+
+$$\int \frac{1}{\sin x (1+\cos x)} dx$$
+
+解：
+
+$$\begin{align*}
+\int \frac{1}{\sin x (1+\cos x)} dx &= \int \frac{1-\cos x}{\sin^3 x} dx \\
+&= \int \csc^3 x dx - \int \csc^2 x \cot x dx \\
+&= -\frac{1}{2}\csc x \cot x + \frac{1}{2}\ln|\tan\frac{x}{2}| + \frac{1}{2}\cot^2 x + C
+\end{align*}$$
+9. 高次三角幂积分
+【9.1】
+
+$$\int \sin^4 x dx$$
+
+解：
+
+$$\begin{align*}
+\int \sin^4 x dx &= \int \left(\frac{1-\cos 2x}{2}\right)^2 dx \\
+&= \frac{1}{4}\int (1 - 2\cos 2x + \cos^2 2x) dx \\
+&= \frac{1}{4}\int \left(1 - 2\cos 2x + \frac{1+\cos 4x}{2}\right) dx \\
+&= \frac{3}{8}x - \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x + C
+\end{align*}$$
+【9.2】
+
+$$\int \frac{1}{1+\sin^4 x} dx$$
+
+解：
+$$
+\begin{align*}
+\int \frac{1}{1+\sin^4 x} dx &= \int \frac{\sec^4 x}{\sec^4 x + \tan^4 x} dx \\
+&= \int \frac{1+\tan^2 x}{1+2\tan^4 x} d\tan x \\
+&= \frac{1}{2\sqrt{2}}\arctan\left(\sqrt{2}\tan x\right) + \frac{1}{2\sqrt{2}}\ln\left|\frac{\sqrt{2}\tan x - 1}{\sqrt{2}\tan x + 1}\right| + C
+\end{align*}$$
+【9.3】
+
+$$\int \cos^5 x dx$$
+
+解：
+
+$$\begin{align*}
+\int \cos^5 x dx &= \int \cos^4 x \cos x dx \\
+&= \int (1-\sin^2 x)^2 d\sin x \\
+&= \sin x - \frac{2}{3}\sin^3 x + \frac{1}{5}\sin^5 x + C
+\end{align*}$$
+10. 其他积分
+【10.1】
+
+$$\int \frac{1}{\sqrt{1-x^2} + \sqrt{1+x^2}} dx$$
+
+解：
+
+$$\begin{align*}
+\int \frac{1}{\sqrt{1-x^2} + \sqrt{1+x^2}} dx &= \frac{1}{2}\int \left(\sqrt{1+x^2} - \sqrt{1-x^2}\right) dx \\
+&= \frac{1}{4}\left(x\sqrt{1+x^2} + \ln\left(x+\sqrt{1+x^2}\right) - x\sqrt{1-x^2} + \arcsin x\right) + C
+\end{align*}$$
+【10.2】
+
+$$\int \frac{1}{x^2 - x - 1} dx
+$$
+解：
+
+$$\begin{align*}
+\int \frac{1}{x^2 - x - 1} dx &= \int \frac{1}{\left(x-\frac{1}{2}\right)^2 - \frac{5}{4}} dx \\
+&= \frac{1}{\sqrt{5}}\ln\left|\frac{2x-1-\sqrt{5}}{2x-1+\sqrt{5}}\right| + C
+\end{align*}$$
+【10.3】
+
+$$\int \frac{1}{e^x (1+e^x)} dx$$
+
+解：
+
+$$\begin{align*}
+\int \frac{1}{e^x (1+e^x)} dx &= \int \left(\frac{1}{e^x} - \frac{1}{1+e^x}\right) dx \\
+&= -e^{-x} - \ln(1+e^{-x}) + C
+\end{align*}$$
+11. 杂项积分
+【11.1】
+
+$$\int \frac{1}{1+e^x - e^{-x}} dx
+$$
+解：
+
+$$\begin{align*}
+\int \frac{1}{1+e^x - e^{-x}} dx &= \int \frac{e^x}{e^x + e^{2x} - 1} dx \\
+&= \frac{1}{\sqrt{5}}\ln\left|\frac{e^x + \frac{1-\sqrt{5}}{2}}{e^x + \frac{1+\sqrt{5}}{2}}\right| + C
+\end{align*}$$
+【11.2】
+
+$$\int \frac{1}{x^2 - 3x + 2} dx$$
+
+解：
+
+$$\int \frac{1}{x^2 - 3x + 2} dx = \int \left(\frac{1}{x-2} - \frac{1}{x-1}\right) dx = \ln\left|\frac{x-2}{x-1}\right| + C$$
+【11.3】
+
+$$\int \frac{1}{(x-a)(x-b)(x-c)} dx \quad (a\neq b\neq c)$$
+
+解：
+
+$$\begin{align*}
+\int \frac{1}{(x-a)(x-b)(x-c)} dx &= \frac{1}{(a-b)(a-c)}\ln|x-a| + \frac{1}{(b-a)(b-c)}\ln|x-b| + \frac{1}{(c-a)(c-b)}\ln|x-c| + C
+\end{align*}$$
+12. 根式积分
+【12.1】
+
+$$\int \frac{1}{\sqrt[3]{(x-1)(x+1)^2}} dx$$
+
+解：令 $t = \sqrt[3]{\frac{x+1}{x-1}}$，则 $x = \frac{t^3+1}{t^3-1}$，$dx = \frac{-6t^2}{(t^3-1)^2} dt$
+
+$$\begin{align*}
+\int \frac{1}{\sqrt[3]{(x-1)(x+1)^2}} dx &= \int \frac{1}{\frac{2}{t^3-1} \cdot \left(\frac{2t^3}{t^3-1}\right)^2} \cdot \frac{-6t^2}{(t^3-1)^2} dt \\
+&= -3\int \frac{1}{t} dt \\
+&= -3\ln|t| + C \\
+&= -3\ln\left|\sqrt[3]{\frac{x+1}{x-1}}\right| + C
+\end{align*}$$
+若被积函数中含有 $\sqrt[n_1]{\frac{ax+b}{cx+d}}$,$\sqrt[n_2]{\frac{ax+b}{cx+d}}$,$\dots$,$\sqrt[n_k]{\frac{ax+b}{cx+d}}$ 时，
+可考虑代换 $\boldsymbol{t = \sqrt[n]{\frac{ax+b}{cx+d}}}$；
+其中 n 是 $n_1$,$n_2$,$\dots$,$n_k$ 的最小公倍数。
\ No newline at end of file