求不定积分 不定积分的求法
\u600e\u6837\u6c42\u4e0d\u5b9a\u79ef\u52061\u3001\u76f4\u63a5\u5229\u7528\u79ef\u5206\u516c\u5f0f\u6c42\u51fa\u4e0d\u5b9a\u79ef\u5206\u3002
2\u3001\u901a\u8fc7\u51d1\u5fae\u5206\uff0c\u6700\u540e\u4f9d\u6258\u4e8e\u67d0\u4e2a\u79ef\u5206\u516c\u5f0f\u3002\u8fdb\u800c\u6c42\u5f97\u539f\u4e0d\u5b9a\u79ef\u5206\u3002\u4f8b\u5982
3\u3001\u8fd0\u7528\u94fe\u5f0f\u6cd5\u5219\uff1a
4\u3001\u8fd0\u7528\u5206\u90e8\u79ef\u5206\u6cd5\uff1a\u222budv=uv-\u222bvdu\uff1b\u5c06\u6240\u6c42\u79ef\u5206\u5316\u4e3a\u4e24\u4e2a\u79ef\u5206\u4e4b\u5dee\uff0c\u79ef\u5206\u5bb9\u6613\u8005\u5148\u79ef\u5206\u3002\u5b9e\u9645\u4e0a\u662f\u4e24\u6b21\u79ef\u5206\u3002\u79ef\u5206\u5bb9\u6613\u8005\u9009\u4e3av\uff0c\u6c42\u5bfc\u7b80\u5355\u8005\u9009\u4e3au\u3002\u4f8b\u5b50\uff1a\u222bInx dx\u4e2d\u5e94\u8bbeU=Inx\uff0cV=x\u3002
\u6269\u5c55\u8d44\u6599\uff1a
\u4e00\u3001\u5e38\u7528\u7684\u79ef\u5206\u516c\u5f0f\u6709\uff1a
\u4e8c\u3001\u6c42\u4e0d\u5b9a\u79ef\u5206\u7684\u6ce8\u610f\u4e8b\u9879\uff1a
1\u3001\u5982\u679cf(x)\u5728\u533a\u95f4I\u4e0a\u6709\u539f\u51fd\u6570\uff0c\u5373\u6709\u4e00\u4e2a\u51fd\u6570F(x)\u4f7f\u5bf9\u4efb\u610fx\u2208I\uff0c\u90fd\u6709F'(x)=f(x)\uff0c\u90a3\u4e48\u5bf9\u4efb\u4f55\u5e38\u6570\u663e\u7136\u4e5f\u6709[F(x)+C]'=f(x).\u5373\u5bf9\u4efb\u4f55\u5e38\u6570C\uff0c\u51fd\u6570F(x)+C\u4e5f\u662ff(x)\u7684\u539f\u51fd\u6570\u3002\u8fd9\u8bf4\u660e\u5982\u679cf(x)\u6709\u4e00\u4e2a\u539f\u51fd\u6570,\u90a3\u4e48f(x)\u5c31\u6709\u65e0\u9650\u591a\u4e2a\u539f\u51fd\u6570\u3002
2\u3001\u867d\u7136\u5f88\u591a\u51fd\u6570\u90fd\u53ef\u901a\u8fc7\u5982\u4e0a\u7684\u5404\u79cd\u624b\u6bb5\u8ba1\u7b97\u5176\u4e0d\u5b9a\u79ef\u5206\uff0c\u4f46\u8fd9\u5e76\u4e0d\u610f\u5473\u7740\u6240\u6709\u7684\u51fd\u6570\u7684\u539f\u51fd\u6570\u90fd\u53ef\u4ee5\u8868\u793a\u6210\u521d\u7b49\u51fd\u6570\u7684\u6709\u9650\u6b21\u590d\u5408\uff0c\u539f\u51fd\u6570\u4e0d\u53ef\u4ee5\u8868\u793a\u6210\u521d\u7b49\u51fd\u6570\u7684\u6709\u9650\u6b21\u590d\u5408\u7684\u51fd\u6570\u79f0\u4e3a\u4e0d\u53ef\u79ef\u51fd\u6570\u3002
\u7b54\u6848\u662f(2/\u221a3)arctan[(2x+1)/\u221a3] + C\uff0c\u7b2c\u4e8c\u7c7b\u6362\u5143\u79ef\u5206\u6cd5\u4f60\u5b66\u4e86\u5417\uff1f
x²+x+1 = (x²+x+1/4)+(3/4) = (x)²+2(x)(1/2)+(1/2)²+(3/4) = (x+1/2)²+(\u221a3/2)²
\u222b dx/(x²+x+1) = \u222b dx/[(x+1/2)²+(\u221a3/2)²]
Let u = x+1/2\uff0cdu = dx
=> \u222b du/[u²+(\u221a3/2)²]
Let u = (\u221a3/2)tanz\uff0cdu = (\u221a3/2)sec²zdz
=> \u222b (\u221a3/2)sec²zdz / [(\u221a3/2)²tan²z+(\u221a3/2)²]
= \u222b (\u221a3/2)sec²zdz / [(\u221a3/2)²(tan²z+1)]
= \u222b sec²zdz / [(\u221a3/2)sec²z] <= \u6052\u7b49\u5f0f1+tan²x = sec²x
= [1/(\u221a3/2)]\u222b dz
= (2/\u221a3)z + C
= (2/\u221a3)arctan[u/(\u221a3/2)] + C
= (2/\u221a3)arctan[(x+1/2)*(2/\u221a3)] + C
= (2/\u221a3)arctan[(2x+1)/\u221a3] + C
∫√x^7dx/x^5
=∫x^(7/2)dx/x^5=∫x^(7/2-5)dx,
=∫x^(-3/2)dx=-2x^(-1/2)+C.
本题用到幂函数的求导公式，逆用即可求出不定积分。
∫dx/(x^2+x)
=∫dx/x(x+1)
=∫dx/x-∫dx/(x+1)
=ln|x|-ln|x+1|+c
=ln|x/(x+1)|+c.
本题主要思路是裂项不定积分，同时用到对数函数求导公式。
1、记t=sinx, 则原式=Ssin^2/cosxdsinx=Ssin^2dx=1/4 S(1-cos2x)d2x=(2x-sin2x)/4+C=(2arcsint-2t根号(1-t^2))/4+C.
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绛旓細鈭垰x^7dx/x^5 =鈭玿^(7/2)dx/x^5=鈭玿^(7/2-5)dx,=鈭玿^(-3/2)dx=-2x^(-1/2)+C.鏈鐢ㄥ埌骞傚嚱鏁扮殑姹傚鍏紡锛岄嗙敤鍗冲彲姹傚嚭涓嶅畾绉垎銆傗埆dx/(x^2+x)=鈭玠x/x(x+1)=鈭玠x/x-鈭玠x/(x+1)=ln|x|-ln|x+1|+c =ln|x/(x+1)|+c.鏈涓昏鎬濊矾鏄椤逛笉瀹氱Н鍒嗭紝鍚屾椂...
绛旓細浠=sint x锛0鈫1锛屽垯t锛0鈫捪/2 鈭玔0锛1]鈭(1-x²)dx =鈭玔0锛毾/2]鈭(1-sin²t)d(sint)=鈭玔0锛毾/2]cos²tdt =½鈭玔0锛毾/2](1+cos2t)dt =(½t+¼sin2t)|[0锛毾/2]=[½路(蟺/2)+¼sin蟺]-(½路0+¼...
