前置知识: 高等数学

不定积分典型例题

11 minIntermediate2026/6/14

不定积分15道典型例题:凑微分、换元积分、分部积分、有理函数积分等核心题型。

1. 凑微分法

例1:求 ∫dxx(1+ln⁡x)\int \frac{dx}{x(1+\ln x)}

解:

∫dxx(1+ln⁡x)=∫d(ln⁡x)1+ln⁡x=∫d(1+ln⁡x)1+ln⁡x=ln⁡∣1+ln⁡x∣+C\int \frac{dx}{x(1+\ln x)} = \int \frac{d(\ln x)}{1+\ln x} = \int \frac{d(1+\ln x)}{1+\ln x} = \ln|1+\ln x| + C


例2:求 ∫dxx(1+x)\int \frac{dx}{\sqrt{x}(1+x)}

解:

∫dxx(1+x)=2∫d(x)1+(x)2=2arctan⁡x+C\int \frac{dx}{\sqrt{x}(1+x)} = 2\int \frac{d(\sqrt{x})}{1+(\sqrt{x})^2} = 2\arctan\sqrt{x} + C


例3:求 ∫cos⁡xsin⁡x+cos⁡xdx\int \frac{\cos x}{\sin x + \cos x} dx

解:

∫cos⁡xsin⁡x+cos⁡xdx=12∫(cos⁡x+sin⁡x)+(cos⁡x−sin⁡x)sin⁡x+cos⁡xdx\int \frac{\cos x}{\sin x + \cos x} dx = \frac{1}{2}\int \frac{(\cos x + \sin x) + (\cos x - \sin x)}{\sin x + \cos x} dx

=12∫dx+12∫d(sin⁡x+cos⁡x)sin⁡x+cos⁡x=x2+12ln⁡∣sin⁡x+cos⁡x∣+C= \frac{1}{2}\int dx + \frac{1}{2}\int \frac{d(\sin x + \cos x)}{\sin x + \cos x} = \frac{x}{2} + \frac{1}{2}\ln|\sin x + \cos x| + C

2. 第二换元法

例4:求 ∫dx(1+x2)2\int \frac{dx}{(1+x^2)^2}

解:令 x=tan⁡tx = \tan t,dx=sec⁡2t dtdx = \sec^2 t\, dt

∫sec⁡2t dtsec⁡4t=∫cos⁡2t dt=12∫(1+cos⁡2t) dt=t2+sin⁡2t4+C\int \frac{\sec^2 t\, dt}{\sec^4 t} = \int \cos^2 t\, dt = \frac{1}{2}\int (1+\cos 2t)\, dt = \frac{t}{2} + \frac{\sin 2t}{4} + C

回代:t=arctan⁡xt = \arctan x

=12arctan⁡x+x2(1+x2)+C= \frac{1}{2}\arctan x + \frac{x}{2(1+x^2)} + C


例5:求 ∫a2−x2 dx\int \sqrt{a^2 - x^2}\, dx(a>0a > 0)

解:令 x=asin⁡tx = a\sin t,dx=acos⁡t dtdx = a\cos t\, dt

∫acos⁡t⋅acos⁡t dt=a2∫cos⁡2t dt=a22(t+sin⁡2t2)+C\int a\cos t \cdot a\cos t\, dt = a^2\int \cos^2 t\, dt = \frac{a^2}{2}\left(t + \frac{\sin 2t}{2}\right) + C

=a22arcsin⁡xa+x2a2−x2+C= \frac{a^2}{2}\arcsin\frac{x}{a} + \frac{x}{2}\sqrt{a^2-x^2} + C

3. 分部积分法

例6:求 ∫xexdx\int x e^x dx

解:设 u=xu = x,dv=exdxdv = e^x dx

∫xexdx=xex−∫exdx=xex−ex+C=(x−1)ex+C\int x e^x dx = x e^x - \int e^x dx = x e^x - e^x + C = (x-1)e^x + C


例7:求 ∫exsin⁡x dx\int e^x \sin x\, dx

解:设 u=sin⁡xu = \sin x,dv=exdxdv = e^x dx

I=exsin⁡x−∫excos⁡x dx=exsin⁡x−excos⁡x−∫exsin⁡x dxI = e^x \sin x - \int e^x \cos x\, dx = e^x \sin x - e^x \cos x - \int e^x \sin x\, dx

I=exsin⁡x−excos⁡x−II = e^x \sin x - e^x \cos x - I

2I=ex(sin⁡x−cos⁡x)2I = e^x(\sin x - \cos x)

I=ex(sin⁡x−cos⁡x)2+CI = \frac{e^x(\sin x - \cos x)}{2} + C


例8:求 ∫ln⁡x dx\int \ln x\, dx

解:设 u=ln⁡xu = \ln x,dv=dxdv = dx

∫ln⁡x dx=xln⁡x−∫x⋅1x dx=xln⁡x−x+C\int \ln x\, dx = x\ln x - \int x \cdot \frac{1}{x}\, dx = x\ln x - x + C


例9:建立递推公式求 In=∫dx(x2+a2)nI_n = \int \frac{dx}{(x^2+a^2)^n}

解:

In=x(x2+a2)n+2n∫x2(x2+a2)n+1dxI_n = \frac{x}{(x^2+a^2)^n} + 2n\int \frac{x^2}{(x^2+a^2)^{n+1}} dx

=x(x2+a2)n+2n∫x2+a2−a2(x2+a2)n+1dx= \frac{x}{(x^2+a^2)^n} + 2n\int \frac{x^2+a^2-a^2}{(x^2+a^2)^{n+1}} dx

=x(x2+a2)n+2nIn−2na2In+1= \frac{x}{(x^2+a^2)^n} + 2nI_n - 2na^2 I_{n+1}

In+1=x2na2(x2+a2)n+2n−12na2InI_{n+1} = \frac{x}{2na^2(x^2+a^2)^n} + \frac{2n-1}{2na^2}I_n

4. 有理函数积分

例10:求 ∫x+3x2−5x+6dx\int \frac{x+3}{x^2-5x+6} dx

解:部分分式分解:

x+3(x−2)(x−3)=Ax−2+Bx−3\frac{x+3}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}

x+3=A(x−3)+B(x−2)x+3 = A(x-3) + B(x-2)

令 x=2x=2:5=−A5 = -A,A=−5A = -5

令 x=3x=3:6=B6 = B,B=6B = 6

∫x+3x2−5x+6dx=−5∫dxx−2+6∫dxx−3=−5ln⁡∣x−2∣+6ln⁡∣x−3∣+C\int \frac{x+3}{x^2-5x+6} dx = -5\int \frac{dx}{x-2} + 6\int \frac{dx}{x-3} = -5\ln|x-2| + 6\ln|x-3| + C


例11:求 ∫dxx3+1\int \frac{dx}{x^3+1}

解:x3+1=(x+1)(x2−x+1)x^3 + 1 = (x+1)(x^2-x+1)

1x3+1=Ax+1+Bx+Cx2−x+1\frac{1}{x^3+1} = \frac{A}{x+1} + \frac{Bx+C}{x^2-x+1}

1=A(x2−x+1)+(Bx+C)(x+1)1 = A(x^2-x+1) + (Bx+C)(x+1)

令 x=−1x=-1:1=3A1 = 3A,A=13A = \frac{1}{3}

比较 x2x^2 系数:0=A+B0 = A + B,B=−13B = -\frac{1}{3}

比较常数项:1=A+C1 = A + C,C=23C = \frac{2}{3}

∫dxx3+1=13ln⁡∣x+1∣−13∫x−2x2−x+1dx\int \frac{dx}{x^3+1} = \frac{1}{3}\ln|x+1| - \frac{1}{3}\int \frac{x-2}{x^2-x+1} dx

=13ln⁡∣x+1∣−16ln⁡(x2−x+1)+13arctan⁡2x−13+C= \frac{1}{3}\ln|x+1| - \frac{1}{6}\ln(x^2-x+1) + \frac{1}{\sqrt{3}}\arctan\frac{2x-1}{\sqrt{3}} + C

5. 三角函数积分

例12:求 ∫sin⁡3x dx\int \sin^3 x\, dx

解:

∫sin⁡3x dx=∫sin⁡x(1−cos⁡2x) dx=−∫(1−cos⁡2x) d(cos⁡x)\int \sin^3 x\, dx = \int \sin x(1-\cos^2 x)\, dx = -\int (1-\cos^2 x)\, d(\cos x)

=−cos⁡x+cos⁡3x3+C= -\cos x + \frac{\cos^3 x}{3} + C


例13:求 ∫sin⁡2xcos⁡2x dx\int \sin^2 x \cos^2 x\, dx

解:

∫sin⁡2xcos⁡2x dx=14∫sin⁡22x dx=14∫1−cos⁡4x2 dx=x8−sin⁡4x32+C\int \sin^2 x \cos^2 x\, dx = \frac{1}{4}\int \sin^2 2x\, dx = \frac{1}{4}\int \frac{1-\cos 4x}{2}\, dx = \frac{x}{8} - \frac{\sin 4x}{32} + C

6. 综合题型

例14:求 ∫dxxx2−1\int \frac{dx}{x\sqrt{x^2-1}}(x>1x > 1)

解法一:令 x=sec⁡tx = \sec t,dx=sec⁡ttan⁡t dtdx = \sec t \tan t\, dt

∫sec⁡ttan⁡tsec⁡t⋅tan⁡t dt=∫dt=t+C=arcsec x+C\int \frac{\sec t \tan t}{\sec t \cdot \tan t}\, dt = \int dt = t + C = \text{arcsec}\, x + C

解法二:令 t=x2−1t = \sqrt{x^2-1}

∫dxxx2−1=∫1x2⋅x dxx2−1=∫dt1+t2=arctan⁡x2−1+C\int \frac{dx}{x\sqrt{x^2-1}} = \int \frac{1}{x^2} \cdot \frac{x\, dx}{\sqrt{x^2-1}} = \int \frac{dt}{1+t^2} = \arctan\sqrt{x^2-1} + C


例15:求 ∫max⁡(1,x2)dx\int \max(1, x^2) dx

解:

max⁡(1,x2)={x2∣x∣≥11∣x∣<1\max(1, x^2) = \begin{cases} x^2 & |x| \geq 1 \\ 1 & |x| < 1 \end{cases}

当 x<−1x < -1:∫x2dx=x33+C1\int x^2 dx = \frac{x^3}{3} + C_1

当 −1≤x<1-1 \leq x < 1:∫1 dx=x+C2\int 1\, dx = x + C_2

当 x≥1x \geq 1:∫x2dx=x33+C3\int x^2 dx = \frac{x^3}{3} + C_3

由连续性:在 x=−1x=-1 处 −13+C1=−1+C2\frac{-1}{3} + C_1 = -1 + C_2

在 x=1x=1 处 1+C2=13+C31 + C_2 = \frac{1}{3} + C_3

取 C2=0C_2 = 0,则 C1=−23C_1 = -\frac{2}{3},C3=23C_3 = \frac{2}{3}

∫max⁡(1,x2)dx={x33−23x<−1x−1≤x<1x33+23x≥1\int \max(1, x^2) dx = \begin{cases} \frac{x^3}{3} - \frac{2}{3} & x < -1 \\ x & -1 \leq x < 1 \\ \frac{x^3}{3} + \frac{2}{3} & x \geq 1 \end{cases}