前置知识：直接积分法

对于函数 f ( x ) , g ( x ) f(x),g(x) f(x),g(x)根据复合函数的求导法则， d d x [ f ( g ( x ) ) ] = f ′ ( g ( x ) ) ⋅ g ′ ( x ) \dfrac{d}{dx}[f(g(x))]=f'(g(x))\cdot g'(x) dxd​[f(g(x))]=f′(g(x))⋅g′(x)

于是 ∫ f ′ ( g ( x ) ) ⋅ g ′ ( x ) d x = f ( g ( x ) ) + C \int f'(g(x))\cdot g'(x)dx=f(g(x))+C ∫f′(g(x))⋅g′(x)dx=f(g(x))+C

令 u = g ( x ) u=g(x) u=g(x)，因为 ∫ f ′ ( u ) d x = f ( u ) + C \int f'(u)dx=f(u)+C ∫f′(u)dx=f(u)+C

则在求 ∫ f ′ ( g ( x ) ) ⋅ g ′ ( x ) \int f'(g(x))\cdot g'(x) ∫f′(g(x))⋅g′(x)时，可以将其变换为 ∫ f ′ ( u ) d u \int f'(u)du ∫f′(u)du，求出结果 f ( u ) + C f(u)+C f(u)+C后再将 u = g ( x ) u=g(x) u=g(x)代入： ∫ f ′ ( g ( x ) ) ⋅ g ′ ( x ) d x = ∫ f ′ ( u ) d u = f ( u ) + C = f ( g ( x ) ) + C \int f'(g(x))\cdot g'(x)dx=\int f'(u)du=f(u)+C=f(g(x))+C ∫f′(g(x))⋅g′(x)dx=∫f′(u)du=f(u)+C=f(g(x))+C

这里涉及到一阶微分的形式不变性。

以上就是第一类换元法，又称凑微分法。

∫ f ( a x + b ) d x = 1 a f ( a x + b ) d ( a x + b ) \int f(ax+b)dx=\dfrac 1af(ax+b)d(ax+b) ∫f(ax+b)dx=a1​f(ax+b)d(ax+b)

∫ f ( a x n + b ) x n − 1 d x = 1 n a ∫ f ( a x n + b ) d ( a x n + b ) \int f(ax^n+b)x^{n-1}dx=\dfrac{1}{na}\int f(ax^n+b)d(ax^n+b) ∫f(axn+b)xn−1dx=na1​∫f(axn+b)d(axn+b)

∫ f ( 1 x ) 1 x 2 d x = − ∫ f ( 1 x ) d ( 1 x ) \int f(\dfrac 1x)\dfrac{1}{x^2}dx=-\int f(\dfrac 1x)d(\dfrac 1x) ∫f(x1​)x21​dx=−∫f(x1​)d(x1​)

∫ f ( x ) 1 x d x = 2 ∫ f ( x ) d ( x ) \int f(\sqrt x)\dfrac{1}{\sqrt x}dx=2\int f(\sqrt x)d(\sqrt x) ∫f(x ​)x ​1​dx=2∫f(x ​)d(x ​)

∫ f ( ln ⁡ x ) 1 x d x = ∫ f ( ln ⁡ x ) d ( ln ⁡ x ) \int f(\ln x)\dfrac 1xdx=\int f(\ln x)d(\ln x) ∫f(lnx)x1​dx=∫f(lnx)d(lnx)

∫ f ( e x ) e x d x = ∫ f ( e x ) d ( e x ) \int f(e^x)e^xdx=\int f(e^x)d(e^x) ∫f(ex)exdx=∫f(ex)d(ex)

∫ f ( sin ⁡ x ) cos ⁡ x d x = ∫ f ( sin ⁡ x ) d ( sin ⁡ x ) \int f(\sin x)\cos xdx=\int f(\sin x)d(\sin x) ∫f(sinx)cosxdx=∫f(sinx)d(sinx)

∫ f ( cos ⁡ x ) sin ⁡ x d x = − ∫ f ( cos ⁡ x ) d ( cos ⁡ x ) \int f(\cos x)\sin xdx=-\int f(\cos x)d(\cos x) ∫f(cosx)sinxdx=−∫f(cosx)d(cosx)

∫ f ( tan ⁡ x ) sec ⁡ 2 x d x = ∫ f ( tan ⁡ x ) d ( tan ⁡ x ) \int f(\tan x)\sec^2 xdx=\int f(\tan x)d(\tan x) ∫f(tanx)sec2xdx=∫f(tanx)d(tanx)

∫ f ( cot ⁡ x ) csc ⁡ 2 x d x = − ∫ f ( cot ⁡ x ) d ( cot ⁡ x ) \int f(\cot x)\csc^2 xdx=-\int f(\cot x)d(\cot x) ∫f(cotx)csc2xdx=−∫f(cotx)d(cotx)

∫ f ( sec ⁡ x ) sec ⁡ x tan ⁡ x d x = ∫ f ( sec ⁡ x ) d ( sec ⁡ x ) \int f(\sec x)\sec x\tan xdx=\int f(\sec x)d(\sec x) ∫f(secx)secxtanxdx=∫f(secx)d(secx)

∫ f ( csc ⁡ x ) csc ⁡ x cot ⁡ x d x = − ∫ f ( csc ⁡ x ) d ( csc ⁡ x ) \int f(\csc x)\csc x\cot xdx=-\int f(\csc x)d(\csc x) ∫f(cscx)cscxcotxdx=−∫f(cscx)d(cscx)

∫ f ( arcsin ⁡ x ) 1 1 − x 2 d x = ∫ f ( arcsin ⁡ x ) d ( arcsin ⁡ x ) \int f(\arcsin x)\dfrac{1}{\sqrt{1-x^2}}dx=\int f(\arcsin x)d(\arcsin x) ∫f(arcsinx)1−x2 ​1​dx=∫f(arcsinx)d(arcsinx)

∫ f ( arctan ⁡ x ) 1 1 + x 2 d x = ∫ f ( arctan ⁡ x ) d ( arctan ⁡ x ) \int f(\arctan x)\dfrac{1}{1+x^2}dx=\int f(\arctan x)d(\arctan x) ∫f(arctanx)1+x21​dx=∫f(arctanx)d(arctanx)

题1： 计算 ∫ 1 2 x + 1 d x \int \dfrac{1}{\sqrt{2x+1}}dx ∫2x+1 ​1​dx

解：原式 = ∫ ( 2 x + 1 ) − 1 2 d x = 1 2 ∫ ( 2 x + 1 ) − 1 2 d ( 2 x + 1 ) = ( 2 x + 1 ) 1 2 + C =\int (2x+1)^{-\frac 12}dx=\dfrac 12\int (2x+1)^{-\frac 12}d(2x+1)=(2x+1)^{\frac 12}+C =∫(2x+1)−21​dx=21​∫(2x+1)−21​d(2x+1)=(2x+1)21​+C

题2： 计算 ∫ x cos ⁡ ( x 2 + 2 ) d x \int x\cos (x^2+2)dx ∫xcos(x2+2)dx

解：原式 = 1 2 ∫ cos ⁡ ( x 2 + 2 ) d ( x 2 + 2 ) = 1 2 sin ⁡ ( x 2 + 2 ) + C =\dfrac 12\int \cos(x^2+2)d(x^2+2)=\dfrac 12\sin(x^2+2)+C =21​∫cos(x2+2)d(x2+2)=21​sin(x2+2)+C

常见凑微分公式并不需要背，只要掌握技巧，这些都是可以推出来的。