大一，高数，求解不定积分，先谢了！
\u5927\u4e00\u9ad8\u6570\u4e0d\u5b9a\u79ef\u5206\u6c42\u89e3\u222b[xe^(arctanx)/(1+x^2)^(3/2)]dx=\u222b[xe^(arctanx)/(1+x^2)^(3/2)]d(arctanx)=\u222b[x/(1+x^2)^(1/2)]d[e^(arctanx)]=[x/(1+x^2)^(1/2)]e^(arctanx)-\u222b[e^(arctanx)/(1+x^2)^(1/2)]-[e^(arctanx)*x^2/(1+x^2)^(3/2)]dx
=[x/(1+x^2)^(1/2)]e^(arctanx)-\u222b[e^(arctanx)*(1+x^2)/(1+x^2)^(3/2)]-[e^(arctanx)*x^2/(1+x^2)^(3/2)]dx =[x/(1+x^2)^(1/2)]e^(arctanx)-\u222b[1/(1+x^2)^(1/2)]d[e^(arctanx)]=[(x-1)/(1+x^2)^(1/2)]e^(arctanx)-\u222b[xe^(arctanx)/(1+x^2)^(3/2)]dx\u2234\u222b[xe^(arctanx)/(1+x^2)^(3/2)]dx=[(x-1)/2(1+x^2)^(1/2)]*e^(arctanx)+C.
\u4ee41+x^4=t\uff0c\u6240\u4ee5\uff1adt/4=x^3dx\uff0c\u539f\u5f0f=(1/4)sdt/(1+t^(1/3)\uff0c\u8fd9\u91cc\u518d\u4f7f\u7528\u516c\u5f0f\uff1a\u4e8c\u9879\u5fae\u5206\u5f0f\uff1a \u222b[(x^m)(a+b*x^n)^p]dx(m,n\u548cp\u4e3a\u6709\u7406\u6570)\uff0c\u7531\u5951\u6bd4\u534f\u592b\u5b9a\u7406\uff0c\u88ab\u79ef\u51fd\u6570\u53ef\u5316\u4e3a\u6709\u7406\u51fd\u6570\u76843\u79cd\u60c5\u51b5\uff1a
\u4e00\u3002p\u4e3a\u6574\u6570\uff0c\u5047\u5b9ax=z^N\uff0c\u5176\u4e2dN\u4e3a\u5206\u6570m\u548cn\u7684\u516c\u5206\u6bcd\uff1b
\u4e8c\u3002(m+1)/n\u4e3a\u6574\u6570\uff0c\u5047\u5b9aa+bx^n=z^N\uff0c\u5176\u4e2dN\u662f\u5206\u6570p\u7684\u5206\u6bcd\uff1b
\u4e09\u3002[(m+1)/n]+p\u4e3a\u6574\u6570\uff0c\u5229\u7528\u4ee3\u6362\uff1a[ax^(-n)]+b=z^N\uff0c\u5176\u4e2dN\u4e3a\u5206\u6570p\u7684\u5206\u6bcd\u3002 \u8bf4\u660e\uff1a\u4e00\u4e8c\u7684\u5047\u5b9a\u5373\u4e3a\u6240\u4f5c\u7684\u4ee3\u6362\u3002\u5bf9\u4e8e\u4e0d\u662f\u4e8c\u9879\u5fae\u5206\u5f0f\u7684\uff0c\u5fc5\u987b\u5316\u5230\u4e8c\u9879\u5fae\u5206\u5f0f\u3002
=1/sin^2 x - 1/ cos^2 x
分别积分
=-cotx-tanx+C
=∫4cos(2x) *dx/(2sinx*cosx)^2
=2*∫cos(2x) *d(2x)/[sin(2x)]^2
=2*∫d[sin(2x)]/[sin(2x)]^2
=2*[-1/sin(2x)] + C
=-2*csc(2x) + C
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绛旓細(4x+1)^10 dx = 1/4*(4x+1)^10 d(4x-1) = 1/44*(4x+1)^11 + C 鈭 lnx/x² dx,棣栧厛灏1/x²鎺ㄨ繘d閲,杩欐槸绉垎杩囩▼= 鈭 lnx d(- 1/x),鐒跺悗浜掕皟鍑芥暟浣嶇疆= - (lnx)/x + 鈭 1/x d(lnx),灏唋nx浠巇閲屾媺鍑烘潵,杩欐槸寰垎杩囩▼= - (lnx)/x + 鈭 1/x * 1/...
