求不定积分 求不定积分:∫xexdx
\u6c42\u4e0d\u5b9a\u79ef\u5206\u88ab\u79ef\u51fd\u6570\u5b9a\u4e49\u57df\u4e3a0\u2264x\u22641
\u4ee4\u221ax=t,\u5219x=t²,dx=2tdt
\u539f\u5f0f=2\u222btarcsintdt
\u67e5\u8868\u5f97\u539f\u5f0f=2*[(t²/2-1/4)arcsint+t\u221a(1-t²)/4+C]
=(x-1/2)arcsin\u221ax+\u221a(x-x²)/2+C
\u5177\u4f53\u56de\u7b54\u5982\u56fe\uff1a
\u6c42\u51fd\u6570f(x)\u7684\u4e0d\u5b9a\u79ef\u5206\uff0c\u5c31\u662f\u8981\u6c42\u51faf(x)\u7684\u6240\u6709\u7684\u539f\u51fd\u6570\uff0c\u7531\u539f\u51fd\u6570\u7684\u6027\u8d28\u53ef\u77e5\uff0c\u53ea\u8981\u6c42\u51fa\u51fd\u6570f(x)\u7684\u4e00\u4e2a\u539f\u51fd\u6570\uff0c\u518d\u52a0\u4e0a\u4efb\u610f\u7684\u5e38\u6570C\u5c31\u5f97\u5230\u51fd\u6570f(x)\u7684\u4e0d\u5b9a\u79ef\u5206\u3002
\u628a\u76f4\u89d2\u5750\u6807\u7cfb\u4e0a\u7684\u51fd\u6570\u7684\u56fe\u8c61\u7528\u5e73\u884c\u4e8ey\u8f74\u7684\u76f4\u7ebf\u628a\u5176\u5206\u5272\u6210\u65e0\u6570\u4e2a\u77e9\u5f62\uff0c\u7136\u540e\u628a\u67d0\u4e2a\u533a\u95f4[a,b]\u4e0a\u7684\u77e9\u5f62\u7d2f\u52a0\u8d77\u6765\uff0c\u6240\u5f97\u5230\u7684\u5c31\u662f\u8fd9\u4e2a\u51fd\u6570\u7684\u56fe\u8c61\u5728\u533a\u95f4[a,b]\u7684\u9762\u79ef\u3002\u5b9e\u9645\u4e0a\uff0c\u5b9a\u79ef\u5206\u7684\u4e0a\u4e0b\u9650\u5c31\u662f\u533a\u95f4\u7684\u4e24\u4e2a\u7aef\u70b9a,b\u3002
\u6269\u5c55\u8d44\u6599\uff1a
\u6c42\u4e0d\u5b9a\u79ef\u5206\u7684\u65b9\u6cd5\uff1a
\u7b2c\u4e00\u7c7b\u6362\u5143\u5176\u5b9e\u5c31\u662f\u4e00\u79cd\u62fc\u51d1\uff0c\u5229\u7528f'(x)dx=df(x)\uff1b\u800c\u524d\u9762\u7684\u5269\u4e0b\u7684\u6b63\u597d\u662f\u5173\u4e8ef\uff08x\uff09\u7684\u51fd\u6570\uff0c\u518d\u628af\uff08x\uff09\u770b\u4e3a\u4e00\u4e2a\u6574\u4f53\uff0c\u6c42\u51fa\u6700\u7ec8\u7684\u7ed3\u679c\u3002\uff08\u7528\u6362\u5143\u6cd5\u8bf4\uff0c\u5c31\u662f\u628af\uff08x\uff09\u6362\u4e3at\uff0c\u518d\u6362\u56de\u6765\uff09
\u5206\u90e8\u79ef\u5206\uff0c\u5c31\u90a3\u56fa\u5b9a\u7684\u51e0\u79cd\u7c7b\u578b\uff0c\u65e0\u975e\u5c31\u662f\u4e09\u89d2\u51fd\u6570\u4e58\u4e0ax\uff0c\u6216\u8005\u6307\u6570\u51fd\u6570\u3001\u5bf9\u6570\u51fd\u6570\u4e58\u4e0a\u4e00\u4e2ax\u8fd9\u7c7b\u7684\uff0c\u8bb0\u5fc6\u65b9\u6cd5\u662f\u628a\u5176\u4e2d\u4e00\u90e8\u5206\u5229\u7528\u4e0a\u9762\u63d0\u5230\u7684f\u2018\uff08x\uff09dx=df\uff08x\uff09\u53d8\u5f62\uff0c\u518d\u7528\u222bxdf(x)=f(x)x-\u222bf\uff08x\uff09dx\u8fd9\u6837\u7684\u516c\u5f0f\uff0c\u5f53\u7136x\u53ef\u4ee5\u6362\u6210\u5176\u4ed6g\uff08x\uff09
\u5e38\u7528\u79ef\u5206\u516c\u5f0f\uff1a
1\uff09\u222b0dx=c
2\uff09\u222bx^udx=(x^(u+1))/(u+1)+c
3\uff09\u222b1/xdx=ln|x|+c
4\uff09\u222ba^xdx=(a^x)/lna+c
5\uff09\u222be^xdx=e^x+c
6\uff09\u222bsinxdx=-cosx+c
7\uff09\u222bcosxdx=sinx+c
8\uff09\u222b1/(cosx)^2dx=tanx+c
9\uff09\u222b1/(sinx)^2dx=-cotx+c
10\uff09\u222b1/\u221a\uff081-x^2) dx=arcsinx+c
用换元法。令u=1/x.
积分=-∫ sinu/udu.这是非初等函数积分；
积分就是-Si(u)，将u=1/x代入，所以这道题目答案就是-Si(1/x)
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