用分部积分法求下列定积分 用分部积分法计算下列定积分
\u5982\u4f55\u7528\u5206\u90e8\u79ef\u5206\u6cd5\u6c42\u8fd9\u4e2a\u5b9a\u79ef\u5206\uff1f\u5148\u628a1/(x-1)^2\u8f6c\u6362\u5230d\u540e\u9762\u53bb\uff0c\u5f53\u7136\u8981\u53d8\u6210\u539f\u51fd\u65701/(1-x)\u7684\u5f62\u5f0f\uff0c\u7136\u540e\u5c31\u53ef\u4ee5\u5e94\u7528\u5206\u90e8\u79ef\u5206\u6cd5\u4e86\u3002
1\u3001\u539f\u5f0f=-\u222b(0,1) xd[e^(-x)]
=-[xe^(-x)|(0,1)-\u222b(0,1)e^(-x)dx]
=-[1/e+e^(-x)|(0,1)]
=-[1/e+1/e-1]
=1-2/e
2\u3001\u539f\u5f0f=(1/2)*\u222b(1,e) lnx[d(x^2)]
=(1/2)*[lnx*x^2|(1,e)-\u222b(1,e) xdx]
=(1/2)*[e^2-(x^2/2)|(1,e)]
=(1/2)*[e^2-e^2/2+1/2]
=(1/4)*(e^2+1)
∫(0→π) xsinx dx
= ∫(0→π) x d(- cosx)
= - xcosx：[0→π] + ∫(0→π) cosx dx
= - π(- 1) + sinx：[0→π]
= π
(2)：
∫(0→1) xe^x dx
= ∫(0→1) x d(e^x)
= xe^x：[0→1] - ∫(0→1) e^x dx
= e - e^x：(0→1)
= e - (e - 1)
= 1
(3)：
∫(1→e) x(x - 1)lnx dx
= ∫(1→e) (x^2 - x)lnx dx
= ∫(1→e) lnx d(x^3/3 - x^2/2)
= (x^3/3 - x^2/2)lnx：(1→e) - ∫(1→e) (x^3/3 - x^2/2)(1/x) dx
= (1/3)e^3 - (1/2)e^2 - ∫(1→e) (x^2/3 - x/2) dx
= (1/3)e^3 - (1/2)e^2 - (x^3/9 - x^2/4)：(1→e)
= (1/3)e^3 - (1/2)e^2 - [(e^3/9 - e^2/4) - (1/9 - 1/4)]
= (2/9)e^3 - (1/4)e^2 - 5/36
(4)：
∫(0→1) x^2e^(2x) dx
= (1/2)∫(0→1) x^2 d(e^(2x))
= (1/2)x^2e^(2x)：(0→1) - (1/2)∫(0→1) 2xe^(2x) dx
= (1/2)e^2 - (1/2)∫(0→1) x d(e^(2x))
= (1/2)e^2 - (1/2)xe^(2x)：(0→1) + (1/2)∫(0→1) e^(2x) dx
= (1/2)e^2 - (1/2)e^2 + (1/4)e^(2x)：(0→1)
= (1/4)(e^2 - 1)
绛旓細(1)锛氣埆(0鈫捪) xsinx dx = 鈭(0鈫捪) x d(- cosx)= - xcosx锛歔0鈫捪] + 鈭(0鈫捪) cosx dx = - 蟺(- 1) + sinx锛歔0鈫捪]= 蟺 (2)锛氣埆(0鈫1) xe^x dx = 鈭(0鈫1) x d(e^x)= xe^x锛歔0鈫1] - 鈭(0鈫1) e^x dx = e - e^x锛(0鈫1)= e - ...
绛旓細鈭玸inx.e^x dx =鈭玸inx de^x =sinx.e^x -鈭玞osx.e^x dx =sinx.e^x -鈭玞osx de^x =sinx.e^x -cosx.e^x -鈭玸inx e^x dx 2鈭玸inx.e^x dx=sinx.e^x -cosx.e^x 鈭玸inx.e^x dx=(1/2)[sinx.e^x -cosx.e^x] +C ...
绛旓細瀹氱Н鍒鏈韩鏄竴涓硷紝鎴栬呭彲浠ヨ鏄竴涓‘瀹氱殑鍊硷紙褰撶劧鍙兘鏄敤鏈煡鍏冪礌鏋勬垚鐨勪篃鍙兘灏辨槸涓涓‘瀹氱殑鏁帮級锛屼竴鑸殑鍒嗗竷绉垎鈭紙a,b锛塮(x)dx=af(a)-bf(b)-鈭(a,b)xdf(x),鍏朵腑鈭紙a,b锛夎〃绀轰笂涓嬮檺鍒嗗埆涓篴锛宐銆俤f锛坸锛夋槸瀵筬锛坸锛夋眰x涓闃跺锛屽鏋滄槸澶氬厓鍑芥暟锛岃姹傚垎鍒眰鍋忓鏁帮紝鍗充互x涓...
绛旓細1銆佸師寮=-鈭(0,1) xd[e^(-x)]=-[xe^(-x)|(0,1)-鈭(0,1)e^(-x)dx]=-[1/e+e^(-x)|(0,1)]=-[1/e+1/e-1]=1-2/e 2銆佸師寮=(1/2)*鈭(1,e) lnx[d(x^2)]=(1/2)*[lnx*x^2|(1,e)-鈭(1,e) xdx]=(1/2)*[e^2-(x^2/2)|(1,e)]=(1/2)...
绛旓細鈭0鈫1 xe^-x dx =-鈭(0,1)xde^(-x)=-[xe^(-x)(0,1)-鈭(0,1)e^(-x)]=-[e+e^x(0,1)]=1-2e 鈭(0鈫1/2) arcsin xdx =xarcsinx(0,1/2)-鈭(0鈫1/2)x/鈭(1-x^2)dx =(1/2)(蟺/6)+[鈭(1-x^2)](0,(1/2)=蟺/12+(鈭3/2)-1 ...
绛旓細鈭玿e^xdx=鈭玿de^x=xe^x-鈭玡^xdx=xe^x-e^x+C 浠ｅ叆涓婁笅闄愬緱 e-e^0=e-1
绛旓細濡傚浘锛屾墍绀猴紝
绛旓細鍥炵瓟锛歝osxdx=dsinx
绛旓細濡傚浘锛
绛旓細鈭玿f(x)dx=xF(x)-鈭獸(x)dx=xF(x)-G(x)+C 瑙ｉ杩囩▼濡備笅锛氳嫢宸茬煡f(x)鐨勫師鍑芥暟涓篎(x),F(x)鐨勫師鍑芥暟涓篏(x),鍒欏彲鐢鍒嗛儴绉垎娉曟眰锛氣埆xf(x)dx=xF(x)-鈭獸(x)dx=xF(x)-G(x)+C
