高数求不定积分，有步骤谢谢
\u9ad8\u6570\uff0c\u6c42\u4e0d\u5b9a\u79ef\u5206\uff0c\u6b65\u9aa41/[(x-3)(x-2)^2]=1/[(x-3)(x-2)]-1/(x-2)^2=1/(x-3)-1/(x-2)-1/(x-2)^2
∴原式=∫dx[1/(x-3)-1/(x-2)-1/(x-2)^2]=ln丨(x-3)/(x-2)丨+1/(x-2)+C。供参考。
绛旓細涓嶅畾绉垎鐨勬櫘鍙 涓嶅畾绉垎鐨勫熀鏈叕寮
绛旓細濡傚浘鎵绀猴細
绛旓細-sin2x/4+C銆4棰橈紝浠=鈭歺锛屽垯x=u²锛宒x=2udu锛屽垯绛夊紡鍖栦负鈭玣 ' (u)2udu=u²锛坋^u+1锛+C锛屽垯鎴愮珛2uf ' (u)=銆恥²锛坋^u+1锛+C銆'=2u锛坋^u+1锛+u²e^u锛屾晠f ' (u)=锛坋^u+1锛+ue^u/2銆備袱杈绉垎锛寰楀埌f(u)=u+e^u锛坲+1锛/2+C銆
绛旓細x^2/(x-1)^10 璇存槑锛歺^2=(x-1+1)^2=(x-1)^2+2(x-1)+1 x^2/(x-1)^10 =((x-1)^2+2(x-1)+1)/(x-1)^10 =1/(x-1)^8+(2(x-1)+1)/(x-1)^10 =(x-1)^(-8) +2 (x-1)^(-9) +(x-1)^(-10)鍘熷紡=鈭玔(x-1)^(-8) +2 (x-1)^(-9) +(...
绛旓細鍥炵瓟锛氬師寮=1/2鈭玪n(x^2+1)d(x^2+1) 璁緓^2+1=u 鍘熷紡=1/2鈭玪nudu =1/2u(lnu-1)+C =1/2(x^2+1)[ln(x^2+1)-1]+C
绛旓細绗竴棰樺啓鎴怱in[x]/((2+Cos[x])(1-Cos[x]^2)鐒跺悗璁総 = Cos[x]锛岄偅涔堝彲浠ュ寲鎴1/(2+t)*(1-t^2)锛岃繖鏄釜鏈夌悊绉垎锛鍐欐垚A/(2+t)+B/(1-t)+C/(1+t)姹傚嚭ABC鍚庯紝绉垎鍗冲彲 绗簩棰樺啓鎴怱qrt[Cos[x/2]^2+2Cos[x/2]Sin[x/2]+Sin[x/2]] = Cos[x/2]+Sin[x/2]鐒跺悗...
绛旓細浠も垰x=t,鍒檟=t2,dx=2tdt 鍘熷紡鍙寲涓猴細鈭玸in2(2t)/t*2tdt =鈭2sin2(2t)dt =鈭(1-cos4t)dt =t-1/4sin4t+C 灏嗏垰x=t浠ｅ叆锛屽緱锛氣垰x-1/4sin4鈭歺+C
绛旓細浠=1/x锛寈=1/t锛宒x=-dt/t^2 鍘熷紡=鈭-dt/t^2*t/(1+1/t^2)=-鈭玹dt/(t^2+1)=-(1/2)*鈭玠(t^2+1)/(t^2+1)=-(1/2)*ln(t^2+1)+C =-ln鈭(1/x^2+1)+C锛屽叾涓瑿鏄换鎰忓父鏁
绛旓細鍙傝冧笅鍥剧殑鍑戝井鍒嗘柟娉
绛旓細鈭玿f ' (x)dx=鈭玿df(x)=xf(x)-鈭玣(x)dx =xf(x)-sinx/x+C 鍏朵腑f(x)=銆恠inx/x銆'姹傚嚭浠ｅ叆鍗冲緱銆5棰橈紝鍥犱负sinx/f(x)=銆恆rctan(cosx)+C銆'=-sinx/锛1+cos²x锛夛紝鎵浠(x)=-锛1+cos²x锛夈傚垯鈭玣(x)dx=-鈭紙1+cos²x锛塪x =-鈭愶紙3+cos2x锛/2銆...
