人工智能教程 - 数学基础课程1.1 - 数学分析（一）26-27 部分分式,分部积分,归纳法,递归序列
∫(1x−1+3x+2)dx=ln∣x−1∣+3∣x+2∣+c\int (\frac{1}{x-1}+\frac{3}{x+2})dx=ln|x-1|+3|x+2|+c∫(x−11+x+23)dx=ln∣x−1∣+3∣x+2∣+c
代数问题 Algebra problem Detect"easy" pieces cover-up method 掩盖法 1.4x−1x2+x+2=4x−1(x−1)(x+2)1. \frac{4x-1}{x^2+x+2}= \frac{4x-1}{(x-1)(x+2)}1.x2+x+24x−1=(x−1)(x+2)4x−1 2.=Ax−1+Bx+22.=\frac{A}{x-1}+\frac{B}{x+2}2.=x−1A+x+2B 3.Solve for A & BSolve for A by multi by(x-1)
4x−1x+2=A+Bx+2...(x−1)\frac{4x-1}{x+2}=A+\frac{B}{x+2}...(x-1)x+24x−1=A+x+2B...(x−1) x=1; 4−11+2=A; A=1x=1; \ \ \ \frac{4-1}{1+2}=A; \ \ \ A=1x=1; 1+24−1=A; A=1 方法： 分解因式 Factor the deominator Q 建立等式 set up cover up 分部积分Partial Fraction always work:But maybe solvely
STEPO:LONG DIVISION (带余除数) P(x)Q(x)=quotient +P(x)Q(x)\large \frac{P(x)}{Q(x)}=quotient \ +\frac{P(x)}{Q(x)}Q(x)P(x)=quotient +Q(x)P(x) Degree P <Degree Q Ex: P(x)(x+4)4.(x2+2x+3).(x2+4)3\frac{P(x)}{(x+4)^4.(x^2+2x+3).(x^2+4)^3}(x+4)4.(x2+2x+3).(x2+4)3P(x) degree = 4+2+6=12 set-up: =A1x+2+A2(x+2)2+A3(x+2)3+A4(x+2)4+B0x+C0x2+2x+3+B1x+C1x2+4+B2x+C2(x2+4)2+B3x+C3(x2+4)3=\frac{A_1}{x+2}+\frac{A_2}{(x+2)^2}+\frac{A_3}{(x+2)^3}+\frac{A_4}{(x+2)^4}+\frac{B_0x+C_0}{x^2+2x+3}+\frac{B_1x+C_1}{x^2+4}+\frac{B_2x+C_2}{(x^2+4)^2}+\frac{B_3x+C_3}{(x^2+4)^3}=x+2A1+(x+2)2A2+(x+2)3A3+(x+2)4A4+x2+2x+3B0x+C0+x2+4B1x+C1+(x2+4)2B2x+C2+(x2+4)3B3x+C3 Integration By Parts (uv)’=u’v+uv’ uv’=(uv)’-u’v ∫uv′dv=uv−∫u′vdx\color{red}\int uv'dv = uv-\int u'vdx∫uv′dv=uv−∫u′vdx ∫abuv′dv=uv∣ab−∫abu′vdx\color{red}\int_a^b uv'dv = uv|_a^b- \int_a^b u'vdx∫abuv′dv=uv∣ab−∫abu′vdx Ex1:∫lnxdx\int lnxdx∫lnxdx
u=lnx; u′=1xu=lnx; \ \ \ u'=\frac{1}{x}u=lnx; u′=x1
v=x; v′=1v=x; \ \ \ \ \ v'=1v=x; v′=1
=xlnxuv−∫1x.xdx=\frac{xlnx}{uv}-\int \frac{1}{x}.xdx=uvxlnx−∫x1.xdx
=xlnx−x+c=xlnx -x+c=xlnx−x+c
∫(lnx)2dx\int (lnx)^2dx∫(lnx)2dx
u=(lnx)2; u′=2(lnx)1xu=(lnx)^2; \ \ \ u'=2(lnx)\frac{1}{x}u=(lnx)2; u′=2(lnx)x1
v=x; v′=1v=x; \ \ \ \ \ v'=1v=x; v′=1
=x(lnx)2−∫2(lnx).1xxdx=x(lnx)^2-\int 2(lnx).\frac{1}{x}xdx=x(lnx)2−∫2(lnx).x1xdx
=x(lnx)2−2(xlnx−x)+c=x(lnx)^2-2(xlnx-x)+c=x(lnx)2−2(xlnx−x)+c
∫(lnx)ndx=x(lnx)n−n∫(lnx)n−1.1x.xdx\int (lnx)^ndx = x(lnx)^n-n\int (lnx)^{n-1}.\frac{1}{x}.xdx∫(lnx)ndx=x(lnx)n−n∫(lnx)n−1.x1.xdx
u=(lnx)n; u′=n(lnx)n−1.1xu=(lnx)^n; \ \ \ u'=n(lnx)^{n-1}.\frac{1}{x}u=(lnx)n; u′=n(lnx)n−1.x1
v=x; v′=1v=x; \ \ \ \ \ v'=1v=x; v′=1
Fn(x)=∫(lnx)ndxF_n(x)= \int (lnx)^ndxFn(x)=∫(lnx)ndx
Fn(x)=x(lnx)n−nFn−1(x)F_n(x)=x(lnx)^n-nF_{n-1}(x)Fn(x)=x(lnx)n−nFn−1(x)
n→n−1→n−2→...→1→0n\rightarrow n-1\rightarrow n-2 \rightarrow ... \rightarrow1\rightarrow0n→n−1→n−2→...→1→0
F0(x)=∫(lnx)0dx=xF_0(x)= \int (lnx)^0dx=xF0(x)=∫(lnx)0dx=x
F1(x)=x(lnx)−F0(x)=x(lnx)−x(+c)F_1(x)= x (lnx) - F_0(x)= x (lnx) - x(+c)F1(x)=x(lnx)−F0(x)=x(lnx)−x(+c)
F2(x)=x(lnx)2−2F1(x)=x(lnx)2−2(xlnx−x)+cF_2(x)= x (lnx)^2 - 2F_1(x)=x(lnx)^2 - 2(xlnx-x)+cF2(x)=x(lnx)2−2F1(x)=x(lnx)2−2(xlnx−x)+c
Ex4： ∫xnexdx\int x^n e^xdx∫xnexdx also works for cosx,sinxu=xn; u′=nxn−1u=x^n; \ \ \ u'=nx^{n-1}u=xn; u′=nxn−1
v=ex; v′=exv=e^x; \ \ \ \ \ v'=e^xv=ex; v′=ex
原式=xn.ex−∫nxn−1exdx=x^n.e^x-\int nx^{n-1}e^xdx=xn.ex−∫nxn−1exdx
递归序列 new recurrenceGn(x)=∫xnexdxG_n(x)=\int x^ne^xdxGn(x)=∫xnexdx
Gn(x)=xn.ex−nGn−1(x)\color{red}G_n(x)=x^n.e^x-nG_{n-1}(x)Gn(x)=xn.ex−nGn−1(x)G0(x)=exG_0(x) =e^xG0(x)=ex
G1(x)=xex−G0(x)=xex−exG_1(x)=xe^x-G_0(x)=xe^x-e^xG1(x)=xex−G0(x)=xex−ex
∫xexdx=xex−ex(+c)\int xe^xdx = xe^x-e^x(+c)∫xexdx=xex−ex(+c)
∫lnxdx=xlnx−x+c\int lnxdx =xlnx-x+c∫lnxdx=xlnx−x+c
horizontal
垂直部分 By vertical slicing
作者：KuFun人工智能
