- ∫kdx=kx+C\int k\text{d}x=kx+C∫kdx=kx+C
- ∫xμdx=xμ+1μ+1+C(μ≠1)\int x^{\mu}\text{d}x=\frac{x^{\mu+1}}{\mu+1}+C(\mu\neq 1)∫xμdx=μ+1xμ+1+C(μ=1)
- ∫dxx=ln∣x∣+C\int \frac{\text{d}x}{x}=\text{ln}|x|+C∫xdx=ln∣x∣+C
- ∫dx1+x2=arctan x+C\int \frac{\text{d}x}{1+x^2}=arctan\ x+C∫1+x2dx=arctan x+C
- ∫dx1−x2=arcsin x+C\int \frac{\text{d}x}{\sqrt {1-x^2}}=arcsin\ x+C∫1−x2dx=arcsin x+C
- ∫cos xdx=sin x+C\int cos\ x\text{d}x=sin\ x+C∫cos xdx=sin x+C
- ∫sin xdx=−cos x+C\int sin\ x\text{d}x=-cos\ x+C∫sin xdx=−cos x+C
- ∫dxcos2 x=tan x+C\int \frac{\text{d}x}{cos^2\ x}=tan\ x+C∫cos2 xdx=tan x+C
- ∫dxsin2 x=−cot x+C\int \frac{\text{d}x}{sin^2\ x}=-cot\ x+C∫sin2 xdx=−cot x+C
- ∫sec xtan xdx=sec x+C\int sec\ xtan\ x\text{d}x=sec\ x+C∫sec xtan xdx=sec x+C
- ∫csc xcot xdx=−csc x+C\int csc\ xcot\ x\text{d}x=-csc\ x+C∫csc xcot xdx=−csc x+C
- ∫exdx=ex+C\int e^x\text{d}x=e^x+C∫exdx=ex+C
- ∫axdx=axln a+C\int a^x\text{d}x=\frac{a^x}{\text{ln}\ a}+C∫axdx=ln aax+C
- ∫tan xdx=−ln ∣cos x∣+C\int tan\ x\text{d}x=-\text{ln}\ |cos\ x|+C∫tan xdx=−ln ∣cos x∣+C
- ∫cot xdx=ln ∣sin x∣+C\int cot\ x\text{d}x=\text{ln}\ |sin\ x|+C∫cot xdx=ln ∣sin x∣+C
- ∫dxa2+x2=1aarctan xa+C\int \frac{\text{d}x}{a^2+x^2}=\frac{1}{a}arctan\ \frac{x}{a}+C∫a2+x2dx=a1arctan ax+C
- ∫dxa2−x2=arcsin xa+C\int \frac{\text{d}x}{\sqrt{a^2-x^2}}=arcsin\ \frac{x}{a}+C∫a2−x2dx=arcsin ax+C
- ∫dxa2+x2=ln (x+a2+x2)+C\int \frac{\text{d}x}{\sqrt{a^2+x^2}}=\text{ln}\ (x+\sqrt{a^2+x^2})+C∫a2+x2dx=ln (x+a2+x2)+C
- ∫dxx2−a2=ln ∣x+x2−a2∣+C\int\frac{\text{d}x}{\sqrt{x^2-a^2}}=\text{ln}\ |x+\sqrt{x^2-a^2}|+C∫x2−a2dx=ln ∣x+x2−a2∣+C
第一类换元法:
找到 f(x)=g[φ(x)]φ′(x)f(x)=g[\varphi(x)]\varphi'(x)f(x)=g[φ(x)]φ′(x)
∫f(x)dx=∫g[φ(x)]φ′(x)dx=∫g[φ(x)]dφ(x)=[∫g(u)du]u=φ(x)\int f(x)\text{d}x=\int g[\varphi(x)]\varphi'(x)\text{d}x=\int g[\varphi(x)]\text{d}\varphi(x)=[\int g(u)\text{d}u]_{u=\varphi(x)}∫f(x)dx=∫g[φ(x)]φ′(x)dx=∫g[φ(x)]dφ(x)=[∫g(u)du]u=φ(x)
第二类换元法:
找到 x=φ(t)x=\varphi(t)x=φ(t)
∫f(x)dx=[∫f(φ(t))φ′(t)dt]t=φ−1(x)\int f(x)\text{d}x=[\int f(\varphi(t))\varphi'(t)\text{d}t]_{t=\varphi^{-1}(x)}∫f(x)dx=[∫f(φ(t))φ′(t)dt]t=φ−1(x)
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