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

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

dxx(1+lnx)=d(lnx)1+lnx=d(1+lnx)1+lnx=ln1+lnx+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)}

dxx(1+x)=2d(x)1+(x)2=2arctanx+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 cosxsinx+cosxdx\int \frac{\cos x}{\sin x + \cos x} dx

cosxsinx+cosxdx=12(cosx+sinx)+(cosxsinx)sinx+cosxdx\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

=12dx+12d(sinx+cosx)sinx+cosx=x2+12lnsinx+cosx+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}

:令 x=tantx = \tan tdx=sec2tdtdx = \sec^2 t\, dt

sec2tdtsec4t=cos2tdt=12(1+cos2t)dt=t2+sin2t4+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=arctanxt = \arctan x

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


例5:求 a2x2dx\int \sqrt{a^2 - x^2}\, dxa>0a > 0

:令 x=asintx = a\sin tdx=acostdtdx = a\cos t\, dt

acostacostdt=a2cos2tdt=a22(t+sin2t2)+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

=a22arcsinxa+x2a2x2+C= \frac{a^2}{2}\arcsin\frac{x}{a} + \frac{x}{2}\sqrt{a^2-x^2} + C

3. 分部积分

例6:求 xexdx\int x e^x dx

:设 u=xu = xdv=exdxdv = e^x dx

xexdx=xexexdx=xexex+C=(x1)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 exsinxdx\int e^x \sin x\, dx

:设 u=sinxu = \sin xdv=exdxdv = e^x dx

I=exsinxexcosxdx=exsinxexcosxexsinxdxI = 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=exsinxexcosxII = e^x \sin x - e^x \cos x - I

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

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


例8:求 lnxdx\int \ln x\, dx

:设 u=lnxu = \ln xdv=dxdv = dx

lnxdx=xlnxx1xdx=xlnxx+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}

In=x(x2+a2)n+2nx2(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+2nx2+a2a2(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+2nIn2na2In+1= \frac{x}{(x^2+a^2)^n} + 2nI_n - 2na^2 I_{n+1}

In+1=x2na2(x2+a2)n+2n12na2InI_{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+3x25x+6dx\int \frac{x+3}{x^2-5x+6} dx

:部分分式分解:

x+3(x2)(x3)=Ax2+Bx3\frac{x+3}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}

x+3=A(x3)+B(x2)x+3 = A(x-3) + B(x-2)

x=2x=25=A5 = -AA=5A = -5

x=3x=36=B6 = BB=6B = 6

x+3x25x+6dx=5dxx2+6dxx3=5lnx2+6lnx3+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}

x3+1=(x+1)(x2x+1)x^3 + 1 = (x+1)(x^2-x+1)

1x3+1=Ax+1+Bx+Cx2x+1\frac{1}{x^3+1} = \frac{A}{x+1} + \frac{Bx+C}{x^2-x+1}

1=A(x2x+1)+(Bx+C)(x+1)1 = A(x^2-x+1) + (Bx+C)(x+1)

x=1x=-11=3A1 = 3AA=13A = \frac{1}{3}

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

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

dxx3+1=13lnx+113x2x2x+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

=13lnx+116ln(x2x+1)+13arctan2x13+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 sin3xdx\int \sin^3 x\, dx

sin3xdx=sinx(1cos2x)dx=(1cos2x)d(cosx)\int \sin^3 x\, dx = \int \sin x(1-\cos^2 x)\, dx = -\int (1-\cos^2 x)\, d(\cos x)

=cosx+cos3x3+C= -\cos x + \frac{\cos^3 x}{3} + C


例13:求 sin2xcos2xdx\int \sin^2 x \cos^2 x\, dx

sin2xcos2xdx=14sin22xdx=141cos4x2dx=x8sin4x32+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 dxxx21\int \frac{dx}{x\sqrt{x^2-1}}x>1x > 1

解法一:令 x=sectx = \sec tdx=secttantdtdx = \sec t \tan t\, dt

secttantsecttantdt=dt=t+C=arcsecx+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=x21t = \sqrt{x^2-1}

dxxx21=1x2xdxx21=dt1+t2=arctanx21+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

max(1,x2)={x2x11x<1\max(1, x^2) = \begin{cases} x^2 & |x| \geq 1 \\ 1 & |x| < 1 \end{cases}

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

1x<1-1 \leq x < 11dx=x+C2\int 1\, dx = x + C_2

x1x \geq 1x2dx=x33+C3\int x^2 dx = \frac{x^3}{3} + C_3

由连续性:在 x=1x=-113+C1=1+C2\frac{-1}{3} + C_1 = -1 + C_2

x=1x=11+C2=13+C31 + C_2 = \frac{1}{3} + C_3

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

max(1,x2)dx={x3323x<1x1x<1x33+23x1\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}