利用分部积分法求解不定积分：∫dx/（x³+1），急求详细解答过程！！！详解！ 求不定积分xarcsinxdx 分布积分法不会``求解详细...
\u222b1/(x^2-4x+8)dx\uff0c\u6c42\u4e0d\u5b9a\u79ef\u5206\uff0c\u5199\u51fa\u8be6\u7ec6\u8fc7\u7a0b\uff0c\u8c22\u8c22\u3002\u5177\u4f53\u56de\u7b54\u5982\u4e0b\uff1a
\u222b1/(x^2-4x+8)dx
=[\uff08x-2)^2+4]/d(x-2)
=1/2arctan(x-2)/2+C
\u5206\u90e8\u79ef\u5206\u6cd5\u7684\u5b9e\u8d28\uff1a
\u5c06\u6240\u6c42\u79ef\u5206\u5316\u4e3a\u4e24\u4e2a\u79ef\u5206\u4e4b\u5dee\uff0c\u79ef\u5206\u5bb9\u6613\u8005\u5148\u79ef\u5206\uff0c\u5b9e\u9645\u4e0a\u662f\u4e24\u6b21\u79ef\u5206\u3002
\u6709\u7406\u51fd\u6570\u5206\u4e3a\u6574\u5f0f\uff08\u5373\u591a\u9879\u5f0f\uff09\u548c\u5206\u5f0f\uff08\u5373\u4e24\u4e2a\u591a\u9879\u5f0f\u7684\u5546\uff09\uff0c\u5206\u5f0f\u5206\u4e3a\u771f\u5206\u5f0f\u548c\u5047\u5206\u5f0f\uff0c\u800c\u5047\u5206\u5f0f\u7ecf\u8fc7\u591a\u9879\u5f0f\u9664\u6cd5\u53ef\u4ee5\u8f6c\u5316\u6210\u4e00\u4e2a\u6574\u5f0f\u548c\u4e00\u4e2a\u771f\u5206\u5f0f\u7684\u548c\uff0c\u53ef\u89c1\u95ee\u9898\u8f6c\u5316\u4e3a\u8ba1\u7b97\u771f\u5206\u5f0f\u7684\u79ef\u5206\u3002
\u89e3\uff1a\u222bxarcsinxdx
=1/2*\u222barcsinxdx^2
=1/2*x^2*arcsinx-1/2\u222bx^2darcsinx
=1/2*x^2*arcsinx-1/2\u222bx^2/\u221a(1-x^2)dx
\u4ee4x=sint\uff0c\u90a3\u4e48\uff0c
\u222bx^2/\u221a(1-x^2)dx
=\u222b(sint)^2/costdsint
=\u222b(sint)^2dt
=\u222b(1-cos2t)/2dt
=1/2t-1/4sin2t+C=1/2t-1/2sint*cost+C
\u53c8x=sint\uff0c\u5219t=arcsinx\uff0ccost=\u221a(1-x^2)\uff0c\u90a3\u4e48
\u222bx^2/\u221a(1-x^2)dx=1/2t-1/2sint*cost+C=1/2arcsinx-1/2*x*\u221a(1-x^2)+C
\u90a3\u4e48\u222bxarcsinxdx=1/2*x^2*arcsinx-1/2\u222bx^2/\u221a(1-x^2)dx
=1/2*x^2*arcsinx-1/4arcsinx+1/4*x*\u221a(1-x^2)+C
\u6269\u5c55\u8d44\u6599\uff1a
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\u5e38\u7528\u7684\u5206\u90e8\u79ef\u5206\u7684\u6839\u636e\u7ec4\u6210\u88ab\u79ef\u51fd\u6570\u7684\u57fa\u672c\u51fd\u6570\u7c7b\u578b\uff0c\u5c06\u5206\u90e8\u79ef\u5206\u7684\u987a\u5e8f\u6574\u7406\u4e3a\u53e3\u8bc0\uff1a\u201c\u53cd\u5bf9\u5e42\u6307\u4e09\u201d\u3002\u5206\u522b\u4ee3\u6307\u4e94\u7c7b\u57fa\u672c\u51fd\u6570\uff1a\u53cd\u4e09\u89d2\u51fd\u6570\u3001\u5bf9\u6570\u51fd\u6570\u3001\u5e42\u51fd\u6570\u3001\u6307\u6570\u51fd\u6570\u3001\u4e09\u89d2\u51fd\u6570\u7684\u79ef\u5206\u3002
\u53c2\u8003\u8d44\u6599\u6765\u6e90\uff1a\u767e\u5ea6\u767e\u79d1-\u5206\u90e8\u79ef\u5206\u6cd5
x^3+1=(x+1)(x²-x+1)
所以
1/(x^3+1)=A/(x+1)+(Bx+C)/(x²-x+1)
A(x²-x+1)+(Bx+C)(x+1)=1
(A+B)x²+(B+C-A)x+A+C=1
A+B=0
B+C=A
A+C=1
A=1/3
B= -1/3
C=2/3
所以
∫1/(x^3+1)dx
=∫［1/3*1/(x+1)+(-1/3*x+2/3)/(x²-x+1)］dx
=1/3*ln|x+1|-1/6*∫(2x-4)/(x²-x+1)dx
=1/3*ln|x+1|-1/6*∫(2x-1)/(x²-x+1)dx+1/2*∫1/(x²-x+1)dx
=1/3*ln|x+1|-1/6*ln(x²-x+1)+1/2*∫1/[(x-1/2)²+(√3/2)²]d(x-1/2)
=1/3*ln|x+1|-1/6*ln(x²-x+1)+1/2*√3/2*arctan[(x-1)/(√3/2)]+C
=1/3*ln|x+1|-1/6*ln(x²-x+1)+1/√3*arctan(2x-1)/√3+C
对分母分解因式 然后拆开
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