求解此题不定积分怎么求，用分部积分法 用分部积分法计算不定积分。。求解
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\u5e0c\u671b\u8fc7\u7a0b\u6e05\u6670\u660e\u767d
=∫ arctanx dx - ∫ [arctanx /(1+x^2) ] dx
=∫ arctanx dx - (1/2)[arctanx]^2
=xarctanx -∫ x/(1+x^2) dx - (1/2)[arctanx]^2
=xarctanx -(1/2)ln(1+x^2) - (1/2)[arctanx]^2 + C
绛旓細鈭 [x^2/(1+x^2)] arctanx dx =鈭 arctanx dx - 鈭 [arctanx /(1+x^2) ] dx =鈭 arctanx dx - (1/2)[arctanx]^2 =xarctanx -鈭 x/(1+x^2) dx - (1/2)[arctanx]^2 =xarctanx -(1/2)ln(1+x^2) - (1/2)[arctanx]^2 + C ...
绛旓細鍥犱负锛(1/cosx)'=(sinx/cos²x)鍘熷紡=鈭玿/cosxd(1/cosx) 鍒嗛儴绉垎 =x/cos²x-鈭1/cosxd(x/cosx)=x/cos²x-鈭1/cosx*(cosx+xsinx/cos²x)dx =x/cos²x-鈭1/cos²xdx-鈭玿sinx/cos³xdx 浠も埆xsinx/cos³xdx=F 鍒橣=x/cos²x...
绛旓細銆愭眰瑙ｆ濊矾銆1銆佽繍鐢ㄥ垎閮ㄧН鍒嗘硶鍏紡锛屽皢e^x鐪嬫垚v,3^x鐪嬫垚u,鍒檇v=d(e^x),du=3^x ln3 dx 2銆佸悎骞跺悓绫婚」锛堝悓涓琛ㄨ揪寮忥級锛屽洜涓哄乏杈瑰拰鍙宠竟锛岄兘鏈 锛屽悎骞跺悗寰楀埌缁撴灉銆傘愭眰瑙ｈ繃绋嬨戙愭湰棰樼煡璇嗙偣銆1銆佷笉瀹氱Н鍒嗐傝f(x)鍦ㄦ煇鍖洪棿I涓婃湁瀹氫箟锛屽鏋滃瓨鍦ㄥ嚱鏁癋(x)锛屼娇寰楀浜庝换涓x鈭圛锛屾垚绔婩'(x)...
绛旓細鈭玪nxdx=xlnx-x+C銆侰涓哄父鏁般傝В绛旇繃绋嬪涓嬶細鈭玪nxdx =xlnx-鈭玿d(lnx)=xlnx-鈭1dx =xlnx-x+C
绛旓細鍒嗘绉垎娉 鍘熷紡=xarctan鈭歺-鈭玿darctan鈭歺 =xarctan鈭歺-鈭玿/(1+x)dx =xarctan鈭歺-鈭(x+1-1)/(1+x)dx =xarctan鈭歺-鈭玔1-1/(1+x)]dx =xarctan鈭歺-x+ln(1+x)+C
绛旓細鍒嗛儴绉垎娉 鈭2xe^(2x) dx = xe^(2x) - 2鈭玡^(2x)/2 dx = (2x - 1)e^(2x)/2 + C涓銆佺Н鍒嗗叕寮忔硶 鐩存帴鍒╃敤绉垎鍏紡姹傚嚭涓嶅畾绉垎銆 浜屻佹崲鍏冪Н鍒嗘硶 鎹㈠厓绉垎娉曞彲鍒嗕负绗竴绫绘崲鍏冩硶涓庣浜岀被鎹㈠厓娉曘 1銆佺涓绫绘崲鍏冩硶锛堝嵆鍑戝井鍒嗘硶锛 閫氳繃鍑戝井鍒嗭紝鏈鍚庝緷鎵樹簬鏌愪釜绉垎鍏紡銆...
绛旓細璁绉垎鍩熶负 x 鈭(-鈭,+鈭)浠わ細F = (-鈭,+鈭)鈭玡^(-x²)dx 鍚屾牱 F= (-鈭,+鈭)鈭玡^(-y²)dy 鐢变簬x,y鏄簰涓嶇浉鍏崇殑鐨勭Н鍒嗗彉閲,鍥犳锛欶² = (-鈭,+鈭)鈭玡^(-x²)dx * (-鈭,+鈭)鈭玡^(-y²)dy = [D]鈭埆e^(-x²)*dx * e^(-...
绛旓細鈭玿lnxdx=1/2鈭玪nxd(x^2)=1/2[(x^2)lnx-鈭(x^2)dlnx]=1/2[(x^2)lnx-鈭玿dx]=1/2(x^2)lnx-1/4(x^2)+C
绛旓細鍘熷紡=鈭玿(sec²x-1)dx =鈭玿 d(tanx-x)=x(tanx-x)-鈭(tanx-x)dx =xtanx-x²-鈭玹anxdx+鈭玿dx =xtanx-x²+鈭玠(cosx)/cosx+x²/2 =xtanx-x²/2+ln|cosx|+C
