大一高数，不定积分基础题，求解～ 大一高数不定积分问题，求解第九题
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=∫(1+e^x-e^x)/(1+e^x)·dx
=∫1·dx-∫e^x/(1+e^x)·dx
=x-∫1/(1+e^x)·d(1+e^x)
=x-ln(1+e^x)+C
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绛旓細鍒嗛儴绉垎锛氣埆ln(1+x^2)dx =xln(1+x^2)-鈭2x^2/(1+x^2)dx =xln(1+x^2)-[鈭2dx-鈭2/(1+x^2)dx]=xln(1+x^2)-2x+2arctanx+C 3.鈭(xcos2x)dx =(1/2)鈭玿dsin2x =(1/2)xsin2x-(1/2)鈭玸in2xdx =(1/2)xsin2x-(1/4)鈭玸in2xd2x =(1/2)xsin2x+(1/4)cos...
绛旓細8銆佸師寮=鈭玿(sec²x-1)dx =鈭玿sec²xdx-鈭玿dx =鈭玿dtanx-x²/2 =xtanx-鈭玹anxdx-x²/2 =xtanx-鈭玸inx/cosx dx-x²/2 =xtanx-鈭-dcosx/cosx-x²/2 =xtanx+ln|cosx|-x²/2+C 11銆佲埆cos(lnx)dx =xcos(lnx)-鈭玿dcos(lnx)=xcos(ln...
绛旓細(1)d(5X)=5dX锛岀瓑寮忎袱杈瑰悓鏃朵箻浠1/5,鏈塪X=1/5d(5X)(3)d(X^2+1)=2XdX锛岀瓑寮忎袱杈瑰悓鏃朵箻浠1/2,鏈塜dX=1/2d(X^2+1)(5)d(鈭歑-2)=1/2(1/鈭歑)dX锛岀瓑寮忎袱杈瑰悓鏃朵箻浠2锛屾湁dX/鈭歑=2d(鈭歑-2)(7)d(arctan2X)=2/(1+4X^2)dX锛岀瓑寮忎袱杈瑰悓鏃朵箻浠1/2,鏈塪X/(1+4X^2)=1/2d...
绛旓細=[x/(1+x^2)^(1/2)]e^(arctanx)-鈭玔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)-鈭玔1/(1+x^2)^(1/2)]d[e^(arctanx)]=[(x-1)/(1+x^2)^(1/2)]e^(arctanx)-鈭玔xe^...
绛旓細瑙ｏ細鏈鏄笁瑙掑嚱鏁瀹氱Н鍒鐨勭粡鍏搁棶棰橈紝鎺ㄥ杩囩▼濡備笅 浣滃彉閲忕疆鎹 y = x - 蟺/2锛屽垯x = y + 蟺/2锛屽師绉垎寮忓寲涓猴細[0,蟺]鈭玿*(sinx)^n *dx = [-蟺/2, 蟺/2]鈭(y+蟺/2)*(sin(y+蟺/2))^n *dy = [-蟺/2, 蟺/2]鈭珁*(cosy)^n *dy + [-蟺/2, 蟺/2]鈭/2*(...
绛旓細^3*dx =鈭玞osx/(sinx)^3*d(sinx)=-2鈭玞osxd[1/(sinx)^2]=-2cosx/(sinx)^2+2鈭1/(sinx)^2*d(cosx)=-2cosx/(sinx)^2+2鈭1/[1-(cosx)^2]*d(cosx)=-2cosx/(sinx)^2+2*(1/2)ln|(1+cosx)/(1-cosx)|+C =-2cosx/(sinx)^2+ln[(1+cosx)/(1-cosx)]+C ...
绛旓細鏂规硶涓锛氣埆锛1/锛坰inx锛媍osx锛塣2锛絛x 锛濓紙1/2锛夆埆锝1/锛伙紙1/鈭2锛塻inx锛嬶紙1/鈭2锛塩osx锛絕2锝漝x 锛濓紙1/2锛夆埆锝1/锛籹inxcos锛埾/4锛夛紜cosxsin锛埾/4锛夛冀^2锝漝x 锛濓紙1/2锛夆埆锝1/锛籹in锛坸锛嬒/4锛夛冀^2锝漝锛坸锛嬒/4锛夛紳锛嶏紙1/2锛塩ot锛坸锛嬒/4锛夛紜C 鏂规硶浜岋細鈭1 锛濓紙1...
绛旓細鈭1/(1+e^x)路dx =鈭(1+e^x-e^x)/(1+e^x)路dx =鈭1路dx-鈭玡^x/(1+e^x)路dx =x-鈭1/(1+e^x)路d(1+e^x)=x-ln(1+e^x)+C
绛旓細(1)鍘熷紡=x^4/4+x^3/3+x+C (2)鍘熷紡=鈭(x^(11/6)-x^(1/3))dx =6/17*x^(17/6)-3/4*x^(4/3)+C (3)鍘熷紡=鈭玔(1+x²)+x]/x(1+x²)dx =鈭(1/x+1/(1+x²))dx =ln|x|+arctanx+C (4)鍘熷紡=鈭(cosx-sinx)dx =sinx+cosx+C ...
