求解一道不定积分，谢谢 求解一道分段函数的原函数的题，谢谢！
\u6c42\u4e00\u9053\u4e0d\u5b9a\u79ef\u5206,\u8c22\u8c22\u4e86.dx/x8\u6b21\u65b9(1-x2\u6b21\u65b9),\u591a\u8c22\u4e86.dx/x8\u6b21\u65b9(1-x2\u6b21\u65b9)\u7684\u4e0d\u5b9a\u79ef\u5206\u7ed3\u679c\u4e3a-1/(7*x^7)-1/(5*x^5)-1/(3*x^3)-1/x+1/2ln|(x+1)/(x-1)|+C\u3002
\u89e3\uff1a\u222b1/(x^8*(1-x^2))dx
=\u222b(1-x^2+x^2)/(x^8*(1-x^2))dx
=\u222b(1-x^2)/(x^8*(1-x^2))dx+\u222bx^2/(x^8*(1-x^2))dx
=\u222b1/x^8dx+\u222b1/(x^6*(1-x^2))dx
=-1/(7x^7)+\u222b(1-x^2+x^2)/(x^6*(1-x^2))dx
=-1/(7x^7)+\u222b1/x^6dx+\u222b1/(x^4*(1-x^2))dx
=-1/(7*x^7)-1/(5*x^5)+\u222b(1-x^2+x^2)/(x^4*(1-x^2))dx
=-1/(7*x^7)-1/(5*x^5)+\u222b1/x^4dx+\u222b1/(x^2*(1-x^2))dx
=-1/(7*x^7)-1/(5*x^5)-1/(3*x^3)+\u222b(1/x^2+1/(1-x^2))dx
=-1/(7*x^7)-1/(5*x^5)-1/(3*x^3)+\u222b1/x^2dx+\u222b1/(1-x^2)dx
=-1/(7*x^7)-1/(5*x^5)-1/(3*x^3)-1/x+1/2*\u222b(1/(1-x)+1/(1+x))dx
=-1/(7*x^7)-1/(5*x^5)-1/(3*x^3)-1/x+1/2ln|(x+1)/(x-1)|+C
\u6269\u5c55\u8d44\u6599\uff1a
1\u3001\u4e0d\u5b9a\u79ef\u5206\u7684\u8fd0\u7b97\u6cd5\u5219
\uff081\uff09\u51fd\u6570\u7684\u548c\uff08\u5dee\uff09\u7684\u4e0d\u5b9a\u79ef\u5206\u7b49\u4e8e\u5404\u4e2a\u51fd\u6570\u7684\u4e0d\u5b9a\u79ef\u5206\u7684\u548c\uff08\u5dee\uff09\u3002\u5373\uff1a
\u222b[a(x)\u00b1b(x)]dx=\u222ba(x)dx\u00b1\u222bb(x)dx
\uff082\uff09\u6c42\u4e0d\u5b9a\u79ef\u5206\u65f6\uff0c\u88ab\u79ef\u51fd\u6570\u4e2d\u7684\u5e38\u6570\u56e0\u5b50\u53ef\u4ee5\u63d0\u5230\u79ef\u5206\u53f7\u5916\u9762\u6765\u3002\u5373\uff1a
\u222bk*a(x)dx=k*\u222ba(x)dx
2\u3001\u4e0d\u5b9a\u79ef\u5206\u7684\u6c42\u89e3\u65b9\u6cd5
\uff081\uff09\u6362\u5143\u79ef\u5206\u6cd5
\u4f8b\uff1a\u222bsinxcosxdx=\u222bsinxdsinx=1/2sin²x+C
\uff082\uff09\u79ef\u5206\u516c\u5f0f\u6cd5
\u4f8b\uff1a\u222be^xdx=e^x\u3001\u222b1/xdx=ln|x|+C\u3001\u222bcosxdx=sinx+C\u3001\u222bsinxdx=-cosx+C
\u53c2\u8003\u8d44\u6599\u6765\u6e90\uff1a\u767e\u5ea6\u767e\u79d1-\u4e0d\u5b9a\u79ef\u5206
\u8be6\u89c1\u56fe\u7247!
= ∫ sec²z/sec⁴z dz
= ∫ cos²z dz
= ∫ (1 + cos2z)/2 dz
= (1/2)z + (1/2)sinzcosz + C
= (1/2)arctanz + (1/2)[x/√(x² + 1)][1/√(x² + 1)] + C
= (1/2)[arctanx + x/(x² + 1)] + C
∫ (x³ + 1)/(x² + 1)² dx
= ∫ [x(x² + 1 - 1) + 1]/(x² + 1)² dx
= ∫ x/(x² + 1) dx + ∫ (1 - x)/(x² + 1)² dx
= (1/2)∫ 1/(x² + 1) d(x² + 1) + ∫ dx/(x² + 1)² - (1/2)∫ 1/(x² + 1)² d(x² + 1)
= (1/2)ln(x² + 1) + 1/[2(x² + 1)] + ∫ dx/(x² + 1)²，代入上面的结果
= (1/2)ln(x² + 1) + 1/[2(x² + 1)] + (1/2)[arctanx + x/(x² + 1)] + C
= (1/2)[(x + 1)/(x² + 1) + ln(x² + 1) + arctanx] + C
绛旓細=sin2u/4+u/2-cos³u/3+C =((x-1)鈭(2x-x²)+arcsin(x-1))/2-(2x-x²)^(3/2)/3+C
绛旓細鍏堣绠椻埆 dx/(x² + 1)²锛屼护x = tanz锛宒x = sec²z dz ==> 鈭(x² + 1) = secz = 鈭 sec²z/sec⁴z dz = 鈭 cos²z dz = 鈭 (1 + cos2z)/2 dz = (1/2)z + (1/2)sinzcosz + C = (1/2)arctanz + (1/2)[x/鈭...
绛旓細绠鍗曡绠椾竴涓嬪嵆鍙紝绛旀濡傚浘鎵绀
绛旓細浠も垰x = t锛寈 = t²锛宒x = d(t²)鈭 arctan鈭歺 dx = 鈭 arctant d(t²)= t²arctant - 鈭 t²/(1 + t²) dt = t²arctant - 鈭 [(1 + t²) - 1]/(1 + t²) dt = t²arctant - 鈭 dt + 鈭 dt/(1...
绛旓細鍘熷紡=鈭玿sinxdx-鈭玜sinxdx =-鈭玿dcosx-a鈭玸inxdx =-(xcosx-鈭玞osxdx)-a(-cosx)=-xcosx+sinx+acosx+C =锛坅-x锛塩osx+sinx+C
绛旓細绛旓細鈭玸inx(cosx)^5 dx = -鈭 (cosx)^5 d(cosx)=-(1/6)*(cosx)^6+C
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绛旓細(1)锛岃1/[(x-1)(x²+1)=a/(x-1)+(bx+c)銆傝В寰梐=1/2銆乥=c=-1/2銆傗埓鍘熷紡=(1/2)ln涓(x-1)+(1/4)ln(x²+1)+(1/2)arctanx+C銆(2)棰橈紝璁1-x=t锛屸埓鍘熷紡=-鈭(1-2t+t²)dt/t^100=(1/99)/t^99-(2/98)/t^98-(1/97)/t^97+C,鍏朵腑t...
绛旓細鏈嬪弸锛屾棰樺緢绠鍗曪紝璇︾粏杩囩▼濡傚浘rt鎵绀猴紝甯屾湜鑳藉府鍒颁綘瑙ｅ喅闂
绛旓細棣栧厛闄嶅箓 =x(1-cos2x)/2 鍏朵腑锛寈cos2xdx=0.5xd(sin2x)=0.5xsin2x+0.5sin2xdx锛岀Н鍒鍗冲彲
