几道微积分的题目。。
\u51e0\u9053\u5fae\u79ef\u5206\u9898\u76ee\uff081\uff09\u4ee4x=sint,\u56e0x\u5c5e\u4e8e(-1,2),\u6545t\u5728(-pi/2,pi/2)\u5185\uff0c\u4e14dx=costdt
\u222bx^2/\u6839\u53f7(1-x^2)dx
=\u222b(sint)^2/cost\u00d7costdt
=\u222b(sint)^2 dt
=\u222b(1-cos2t)/2 dt
=t/2-sin2t/4+C
=arcsinx/2-x\u00d7\u6839\u53f7(1-x^2)/2+C
(2)\u222bln(1+x)dx
=\u222bln(x+1)d(x+1)
=(x+1)ln(x+1)-\u222b(x+1)dln(x+1)
=(x+1)ln(x+1)-\u222b1 dx
=(x+1)ln(x+1)-x+C
(3)\u222bx*cos\u5e73\u65b9xdx
=\u222bx(1+cos2x)/2 dx
=\u222bx/2 dx+ \u222bxcos2x/2 dx
=x^2/4+x sin2x/4-\u222bsin2x/4 dx
=x^2/4+x sin2x/4+cos2x/8 + C
\u6211\u57fa\u672c\u4e0d\u7ed9\u7b54\u6848\uff0c\u7ed9\u70b9\u60f3\u6cd5\uff0c\u7b54\u6848\u8fd8\u5f97\u60a8\u81ea\u5df1\u6765\u5199\u3002
\u6709\u4e9b\u9898\u6545\u610f\u6ca1\u6709\u6309\u4f60\u8981\u6c42\u7684\u5199\u5f97\u7279\u8be6\u7ec6\uff0c\u8981\u4e0d\u7136\u4f60\u81ea\u5df1\u4e0d\u7422\u78e8\uff0c\u5c31\u66f4\u770b\u4e0d\u61c2\u4e86\u3002
\u53e6\u5916\uff0c\u697c\u4e0a\uff0c\u4e0d\u613f\u610f\u56de\u7b54\u4e3a\u4ec0\u4e48\u4e0d\u7ed5\u884c\uff1f\u975e\u8981\u8dd1\u4e0a\u6765\u62b1\u6028\u51e0\u53e5\uff1f
1\u3001\u628ax\u6362\u6210-1/x\uff0c\u90a3\u4e48-1/x\u5c31\u6362\u6210\u4e86-1/(-1/x)=x\uff0c\u4e5f\u5c31\u662f
af(-1/x)+bf(x)=-sin(1/x)\uff0c
\u8ddf\u539f\u6765\u90a3\u4e2a\u65b9\u7a0b\u653e\u4e00\u8d77\uff0c\u7531a+b\u22600\uff0c\u5c31\u53ef\u4ee5\u6c42\u51faf(x)+f(-1/x)\uff0c\u518d\u7531a\u2260b\u5c31\u80fd\u6c42\u51faf(x)\u3002
2\u3001\u6211\u5c31\u7528a\u548cb\u4e86\uff0c\u5e0c\u814a\u5b57\u6bcd\u6253\u7740\u8d39\u52b2\u3002\u9996\u5148b\u4e0d\u662f0\u3002\u5047\u5982b\u662f\u8d1f\u7684\uff08\u6b63\u7684\u90a3\u79cd\u60c5\u51b5\u5927\u4f53\u4e0a\u7c7b\u4f3c\uff0c\u7ed3\u679c\u4e00\u6837\uff09\uff0c\u90a3\u4e48
(x+1)^b-x^b=(x^(-b)-(x+1)^(-b)) / (x^(-b)(x+1)^(-b))
\u5927\u7ea6\u662fbx^(-b-1)/x^(2b)=bx^(b-1)\u91cf\u7ea7\u7684\uff0c
\u90a3\u4e48\u73b0\u5728\uff0cx^a/(bx^(b-1))\u8d8b\u4e8e8\uff0c\u6240\u4ee5\u5fc5\u987bb=1/8\u5e76\u4e14a=b\u3002
3\u3001\u8fd9\u4e2a\u4f60\u8ba9t=e^x\uff0c\u90a3\u4e481/t=e^(-x)\uff0c\u89e3\u4e2a\u4e8c\u6b21\u65b9\u7a0bt^2-2yt-1=0\uff0c\u7136\u540e\u5728\u4ee3\u8fdbx=lnt\u3002
4\u3001\u8fd9\u4e2a\u5f53\u7136\u4e0d\u5b58\u5728\u3002\u4e24\u4e2a\u5b50\u5217\u6709\u4e0d\u540c\u7684\u6781\u9650\u4e86\u3002\u53cd\u4e4b\u5047\u5982\u8fd9\u4e2a\u6781\u9650\u5b58\u5728\uff0c\u662fC\u7684\u8bdd\uff0c\u4f60\u53ef\u4ee5\u8bc1\u660e\u5b83\u7684\u4efb\u4f55\u5b50\u5217\uff08\u6bd4\u5982x(n)\u548cy(n)\uff09\u7684\u6781\u9650\u90fd\u662fC\uff0c\u90a3\u4e48A=C=B\u5c31\u77db\u76fe\u4e86\u3002
5\u3001x=1\u662f\u53ef\u53bb\u95f4\u65ad\u70b9\uff08\u7b2c\u4e00\u7c7b\uff09\uff0c\u53ef\u4ee5\u76f4\u63a5\u7528l'Hospital\u6cd5\u5219\u6c42\u4e2a\u6781\u9650\uff1b
x\u8d8b\u4e8e0\u6216\u80052\u7684\u65f6\u5019f\u8d8b\u4e8e\u6b63\u65e0\u7a77\uff0c\u662f\u4e2a\u6781\u70b9\uff08\u7b2c\u51e0\u7c7b\u7684\u4f60\u5c31\u81ea\u5df1\u770b\u4e00\u4e0b\u5b9a\u4e49\u597d\u4e86\uff09\u3002
\u5176\u4f59\u70b9\u4e0a\u5206\u5b50\u3001\u5206\u6bcd\u90fd\u6709\u5b9a\u4e49\uff0c\u6240\u4ee5\u4e0d\u662f\u95f4\u65ad\u70b9\u3002
6\u3001\u8fd9\u4e2a\u6ca1\u6709\u6781\u9650\u3002\u4f60\u53ef\u4ee5\u76f4\u63a5\u8bc1\u660e\u5b83\u5728x\u8d8b\u4e8e0\u7684\u65f6\u5019\uff0c\u548c|x|/x\u662f\u7b49\u4ef7\u7684\uff1b\u6216\u8005\u4f60\u53ef\u4ee5\u66f4\u76f4\u63a5\u5730\u8bc1\u660e\uff0c\u5b83\u7684\u53f3\u6781\u9650\u662f1\uff0c\u5de6\u6781\u9650\u662f-1\uff0c\u6240\u4ee5\u6ca1\u6709\u6781\u9650\u3002
2、f(x)=(x³)'=3x²
f '(x)=6x
3、两边求导得：f(x)=-sin2x
4、∫ f '(x) dx=f(x)+C=sinx + C
1、∫f(x)dx=∫x²dx=x³/3，原函数是x³/3+C
2、f(x)的原函数是x³，则f'(x)=2x²
3、∫f(x)dx=cos2x+c，则原函数f(x)=1/2sin2x
4、原函数是f(x)=sinx，则∫f'(x)dx=∫cosxdx=-sinx+C
没有题目啊
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