用不定积分第一类换元法怎么求
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绛旓細1銆佺浜岀被鎹㈠厓绉垎娉 浠=鏍瑰彿涓(x-1)锛屽垯x=t^2+1锛宒x=2tdt 鍘熷紡=鈭(t^2+1)/t*2tdt =2鈭(t^2+1)dt =(2/3)*t^3+2t+C =(2/3)*(x-1)^(3/2)+2鏍瑰彿涓(x-1)+C锛屽叾涓瑿鏄换鎰忓父鏁 2銆绗竴绫绘崲鍏冪Н鍒嗘硶 鍘熷紡=鈭(x-1+1)/鏍瑰彿涓(x-1)dx =鈭玔鏍瑰彿涓(x-1)+1/...
绛旓細du = e^x dx dx = du/u 鈭 dx/[ e^x +e^(-x) ]=鈭 (du/u)/( u +1/u )=鈭 du/( u^2 +1 )=arctanu + C =arctan(e^x) + C
绛旓細鍑戝井鍒嗘硶銆绗竴绫绘崲鍏鍏跺疄灏辨槸涓绉嶆嫾鍑戯紝鍒╃敤f'(x)dx=df(x)锛涜屽墠闈㈢殑鍓╀笅鐨勬濂芥槸鍏充簬f锛坸锛夌殑鍑芥暟锛屽啀鎶奻锛坸锛夌湅涓轰竴涓暣浣擄紝姹傚嚭鏈缁堢殑缁撴灉銆傦紙鐢ㄦ崲鍏冩硶璇达紝灏辨槸鎶奻锛坸锛夋崲涓簍锛屽啀鎹㈠洖鏉ワ級銆涓嶅畾绉垎鐨勫叕寮 1銆佲埆 a dx = ax + C锛宎鍜孋閮芥槸甯告暟 2銆佲埆 x^a dx = [x^(a ...
绛旓細涓嶅畾绉垎鐨勬崲鍏冪Н鍒嗘硶鏂规硶濡備笅锛涓銆佺涓绫绘崲鍏冩硶 锛堝嵆鍑戝井鍒嗘硶锛夐氳繃鍑戝井鍒嗭紝鏈鍚庝緷鎵樹簬鏌愪釜绉垎鍏紡銆傝繘鑰屾眰寰楀師涓嶅畾绉垎銆備簩銆佺浜岀被鎹㈠厓娉 1銆佺浜岀被鎹㈠厓娉曠粡甯哥敤浜庢秷鍘昏绉嚱鏁颁腑鐨勬牴寮忋傚綋琚Н鍑芥暟鏄鏁板緢楂樼殑浜岄」寮忕殑鏃跺欙紝涓轰簡閬垮厤绻佺悙鐨勫睍寮寮忥紝鏈夋椂涔熷彲浠ヤ娇鐢ㄧ浜岀被鎹㈠厓娉曟眰瑙ｃ傚父鐢ㄧ殑鎹...
绛旓細鍏蜂綋鍥炵瓟濡備笅锛氬埄鐢ㄧ浜绉垎鎹㈠厓娉 浠=tanu u鈭堬紙-蟺/2,蟺/2锛夊垯鈭垰锛1+x²锛塪x =鈭玸ec³udu =鈭玸ecudtanu =secutanu-鈭玹anudsecu=secutanu-鈭玹an²usecudu =secutanu 鈭玸ec³udu+鈭玸ecudu=secutanu+1/2ln|secu+tanu|-鈭玸ecudu 鎵浠モ埆sec³udu =1/2锛...
绛旓細绗竴绫绘崲鍏冩硶锛氳f(u)鍏锋湁鍘熷嚱鏁F(U)锛屽嵆銆侳'(U)=f(u)锛屸埆f(u)du=F(U)+C銆傚鏋渦鏄腑闂村彉閲,u=蠁(x)锛屼笖璁鞠(x)鍙井锛岄偅涔堬紝鏍规嵁澶嶅悎鍑芥暟寰垎娉曟湁锛歞F(蠁(x))=f(蠁(x))蠁'(x)dx銆備粠鑰屾牴鎹涓嶅畾绉垎鐨勫畾涔夊氨寰楋細鈭玣[蠁(x)]蠁'(x)dx=F[蠁(x)]+C=[鈭玣(u)du] (u...
绛旓細1銆佺Н鍒嗗叕寮忔硶锛氱洿鎺ュ埄鐢ㄧН鍒嗗叕寮忔眰鍑轰笉瀹氱Н鍒嗐2銆佺涓绫绘崲鍏冩硶锛堝嵆鍑戝井鍒嗘硶锛夛細閫氳繃鍑戝井鍒嗭紝鏈鍚庝緷鎵樹簬鏌愪釜绉垎鍏紡锛岃繘鑰屾眰寰楀師涓嶅畾绉垎銆傜Н鍒嗗父鐢ㄦ硶鍒欏叕寮忥細1銆佲埆0dx=c 涓嶅畾绉垎鐨勫畾涔夈2銆佲埆x^udx=(x^(u+1))/(u+1)+c銆3銆佲埆1/xdx=ln|x|+c銆4銆佲埆a^xdx=(a^x)/lna+c銆5銆...
绛旓細浣犲簲鐢ㄧ殑鏄浜岀被鎹㈠厓娉曘绗竴绫绘崲鍏冩硶浜︾О鍑戝井鍏冩硶銆
绛旓細A=鈭玞osx/(sinx+cosx)dx B=鈭玸inx/(sinx+cosx)dx A+B=鈭紙cosx+sinx锛/(sinx+cosx)dx =鈭玠x =x+c (1) A-B =鈭紙cosx-sinx锛/(sinx+cosx)dx =鈭紙d(cosx+sinx)/(sinx+cosx)=ln(cosx+sinx)+c
绛旓細脳 涓汉銆佷紒涓氱被渚垫潈鎶曡瘔 杩濇硶鏈夊淇℃伅,璇峰湪涓嬫柟閫夋嫨鍚庢彁浜 绫诲埆 鑹叉儏浣庝織 娑夊珜杩濇硶鐘姜 鏃舵斂淇℃伅涓嶅疄 鍨冨溇骞垮憡 浣庤川鐏屾按 鎴戜滑浼氶氳繃娑堟伅銆侀偖绠辩瓑鏂瑰紡灏藉揩灏嗕妇鎶ョ粨鏋滈氱煡鎮ㄣ 璇存槑 0/200 鎻愪氦 鍙栨秷 棰嗗彇濂栧姳 鎴戠殑璐㈠瘜鍊 -- 鍘荤櫥褰 鎴戠殑鐜伴噾 -- 鍘荤櫥褰 鍋氫换鍔″紑瀹濈 绱瀹屾垚 0 涓换鍔 10浠诲姟 鐣ョ暐...
