MIT_单变量微积分_27

深碍√TFBOYSˉ_ 2022-02-20 15:16 219阅读 0赞

### 1.分部积分 ###

**推导：** ( u v ) ′ = u ′ v + u v ′ ( 乘 法 法 则 ) ⇒ u v ′ = ( u v ) ′ − u ′ v ∫ u v ′ = u v − ∫ u ′ v d x , ∫ a b u v ′ d x = u v ∣ a b − ∫ a b u ′ v d x (uv)&\#x27;=u&\#x27;v+uv&\#x27;(乘法法则 )\\Rightarrow uv&\#x27;=(uv)&\#x27;-u&\#x27;v\\\\ \\int uv&\#x27;= uv-\\int u&\#x27;vdx,\\int\_a^buv&\#x27;dx=uv|\_a^b-\\int\_a^bu&\#x27;vdx (uv)′=u′v\+uv′(乘法法则)⇒uv′=(uv)′−u′v∫uv′=uv−∫u′vdx,∫abuv′dx=uv∣ab−∫abu′vdx  
`Ex1:` ∫ l n x d x = l n x ⋅ x − ∫ 1 x ⋅ x d x = x l n x − x + C u = l n x , u ′ = 1 x ; v = x , v ′ = 1 \\int lnxdx=lnx \\cdot x-\\int \\frac\{1\}\{x\} \\cdot xdx=xlnx-x+C\\\\ u=lnx,u&\#x27;=\\frac\{1\}\{x\};v=x,v&\#x27;=1 ∫lnxdx=lnx⋅x−∫x1⋅xdx=xlnx−x\+Cu=lnx,u′=x1;v=x,v′=1  
`Ex2:`  ∫ ( l n x ) 2 d x , u = ( l n x ) 2 , u ′ = 2 ⋅ l n x ⋅ 1 x v = x , v ′ = 1 原 式 = ( l n x ) 2 ⋅ x − ∫ 2 ⋅ l n x ⋅ 1 x ⋅ x d x = ( l n x ) 2 ⋅ x − ∫ 2 ⋅ l n x d x ( 由 E 1 可 知 ) = ( l n x ) 2 x − 2 ( x l n x − x ) + C \\int(lnx)^2dx,u=(lnx)^2,u&\#x27;=2\\cdot lnx\\cdot\\frac\{1\}\{x\}\\\\ v=x,v&\#x27;=1\\\\ 原式=(lnx)^2\\cdot x-\\int2\\cdot lnx\\cdot \\frac\{1\}\{x\} \\cdot xdx\\\\ =(lnx)^2\\cdot x-\\int2 \\cdot lnxdx(由E1可知)\\\\ =(lnx)^2x-2(xlnx-x)+C ∫(lnx)2dx,u=(lnx)2,u′=2⋅lnx⋅x1v=x,v′=1原式=(lnx)2⋅x−∫2⋅lnx⋅x1⋅xdx=(lnx)2⋅x−∫2⋅lnxdx(由E1可知)=(lnx)2x−2(xlnx−x)\+C  
`Ex3:` ∫ ( l n x ) n d x = ( l n x ) n ⋅ x − ∫ n ⋅ ( l n x ) n − 1 ⋅ 1 x ⋅ x d x u = ( l n x ) n , u ′ = n ( l n x ) n − 1 ⋅ 1 x ; v = x , v ′ = 1 原 式 = ( l n x ) n ⋅ x − ∫ n ⋅ ( l n x ) n − 1 d x \\int(lnx)^ndx=(lnx)^n\\cdot x-\\int n \\cdot (lnx)^\{n-1\} \\cdot \\frac\{1\}\{x\} \\cdot xdx\\\\ u=(lnx)^n,u&\#x27;=n(lnx)^\{n-1\}\\cdot \\frac\{1\}\{x\};v=x,v&\#x27;=1\\\\ 原式=(lnx)^n \\cdot x-\\int n \\cdot(lnx)^\{n-1\}dx ∫(lnx)ndx=(lnx)n⋅x−∫n⋅(lnx)n−1⋅x1⋅xdxu=(lnx)n,u′=n(lnx)n−1⋅x1;v=x,v′=1原式=(lnx)n⋅x−∫n⋅(lnx)n−1dx  
**从上式可以总结一个结论:** F n ( x ) = ∫ ( l n x ) n d x , F n ( x ) = x ( l n x ) n − n F n − 1 ( x ) F\_n(x)=\\int(lnx)^ndx,F\_n(x)=x(lnx)^n-nF\_\{n-1\}(x) Fn(x)=∫(lnx)ndx,Fn(x)=x(lnx)n−nFn−1(x)  
**将上式进行倒推：** F 0 ( x ) = ∫ ( l n x ) 0 d x = x ; F 1 ( x ) = x l n x − F 0 ( x ) = x l n x − x + C F 2 ( x ) = x ( l n x ) 2 − 2 F 1 ( x ) = x ( l n x ) 2 − 2 ( x l n x − x ) + C F\_0(x)=\\int (lnx)^0 dx=x;F\_1(x)=xlnx-F\_0(x)=xlnx-x+C\\\\ F\_2(x)=x(lnx)^2-2F\_1(x)=x(lnx)^2-2(xlnx-x)+C F0(x)=∫(lnx)0dx=x;F1(x)=xlnx−F0(x)=xlnx−x\+CF2(x)=x(lnx)2−2F1(x)=x(lnx)2−2(xlnx−x)\+C  
`Ex` ∫ x n e x d x , 令 u = x n , u ′ = n x n − 1 , v = e x , v ′ = e x 原 式 = x n e x − ∫ n x n − 1 e x d x G n ( x ) = ∫ x n e x d x , G n ( x ) = x n ⋅ e x − n G n − 1 ( x ) \\int x^n e^xdx,令u=x^n,u&\#x27;=nx^\{n-1\},v=e^x,v&\#x27;=e^x\\\\ 原式=x^ne^x-\\int nx^\{n-1\}e^xdx\\\\ G\_n(x)=\\int x^ne^xdx,G\_n(x)=x^n \\cdot e^x-nG\_\{n-1\}(x) ∫xnexdx,令u=xn,u′=nxn−1,v=ex,v′=ex原式=xnex−∫nxn−1exdxGn(x)=∫xnexdx,Gn(x)=xn⋅ex−nGn−1(x)