简单的高数，不定积分题目，换元法，求数学帝来帮帮忙！谢了
\u7b80\u5355\u7684\u9ad8\u6570\uff0c\u4e0d\u5b9a\u79ef\u5206\u9898\u76ee\uff0c\u6362\u5143\u6cd5\uff0c\u6c42\u6570\u5b66\u5e1d\u6765\u5e2e\u5e2e\u5fd9\uff01\u8c22\u4e86\u4e09\u89d2\u6362\u5143 \u5efa\u8bae\u60a8\u81ea\u884c\u767e\u5ea6\u2026\u2026
\u6700\u597d\u770b\u8bfe\u672c\u2026\u2026
原式=-∫tdt/√(t^4+1)
=-1/2*∫d(t^2)/√[(t^2)^2+1]
=-1/2*ln|t^2+√(t^4+1)|+C
=-1/2*ln|1/x^2+√(1/x^4+1)|+C
2、令x=sint dx=costdt
原式=∫costdt/(sint+cost)
令A=∫costdt/(sint+cost) B=∫sintdt/(sint+cost)
A+B=∫(sint+cost)dt/(sint+cost)=t+C1
A-B=∫(cost-sint)dt/(sint+cost)=∫d(sint+cost)/(sint+cost)=ln|sint+cost|+C2
A=(t+ln|sint+cost|)/2+C
所以原式=(arcsinx+ln|x+√(1-x^2)|)/2+C
3、原式=∫[1+√x-√(1+x)]dx/(1+2√x+x-1-x)
=1/2*∫x^(-1/2)+1-√(1+x)/√x dx
=√x+x/2-1/2*∫√(1+x)dx/√x
令t=√x t^2=x 1+x=t^2+1 dx=2tdt
原式=√x+x/2-∫√(t^2+1)dt
=√x+x/2-t/2*√(t^2+1)-1/2*ln|t+√(t^2+1)|+C
=√x+x/2-1/2*√(x^2+x)-1/2*ln|√x+√(x+1)|+C
4、令t=lnx x=e^t dx=e^tdt
原式=∫√(1+t)dt/t
再令u=√(1+t) t=u^2-1 dt=2udu
原式=2*∫u^2du/(u^2-1)
=2*∫1+1/(u^2-1) du
=2*∫du+∫du/(u-1)-∫du/(u+1)
=2u+ln|u-1|-ln|u+1|+C
=2√(1+t)+ln|√(1+t)-1|-ln|√(1+t)+1|+C
=2√(1+lnx)+ln|√(1+lnx)-1|-ln|√(1+lnx)+1|+C
一、令x=1/t dx=-dt/t^2
原式=-∫tdt/√(t^4+1)
=-1/2*∫d(t^2)/√[(t^2)^2+1]
=-1/2*ln|t^2+√(t^4+1)|+C
=-1/2*ln|1/x^2+√(1/x^4+1)|+C
二、令x=sint dx=costdt
原式=∫costdt/(sint+cost)
令A=∫costdt/(sint+cost) B=∫sintdt/(sint+cost)
A+B=∫(sint+cost)dt/(sint+cost)=t+C1
A-B=∫(cost-sint)dt/(sint+cost)=∫d(sint+cost)/(sint+cost)=ln|sint+cost|+C2
A=(t+ln|sint+cost|)/2+C
所以原式=(arcsinx+ln|x+√(1-x^2)|)/2+C
三、原式=∫[1+√x-√(1+x)]dx/(1+2√x+x-1-x)
=1/2*∫x^(-1/2)+1-√(1+x)/√x dx
=√x+x/2-1/2*∫√(1+x)dx/√x
令t=√x t^2=x 1+x=t^2+1 dx=2tdt
原式=√x+x/2-∫√(t^2+1)dt
=√x+x/2-t/2*√(t^2+1)-1/2*ln|t+√(t^2+1)|+C
=√x+x/2-1/2*√(x^2+x)-1/2*ln|√x+√(x+1)|+C
四、令t=lnx x=e^t dx=e^tdt
原式=∫√(1+t)dt/t
再令u=√(1+t) t=u^2-1 dt=2udu
原式=2*∫u^2du/(u^2-1)
=2*∫1+1/(u^2-1) du
=2*∫du+∫du/(u-1)-∫du/(u+1)
=2u+ln|u-1|-ln|u+1|+C
=2√(1+t)+ln|√(1+t)-1|-ln|√(1+t)+1|+C
=2√(1+lnx)+ln|√(1+lnx)-1|-ln|√(1+lnx)+1|+C
绛旓細涓嶅畾绉垎鐨勭浜岀被鎹㈠厓娉 绗簩绫鎹㈠厓娉曠殑鐩殑鏄负浜嗘秷鍘绘牴鍙凤紝鍖栦负绠鍗鍑藉紡鐨勪笉瀹氱Н鍒嗐傚畠鍒嗕负鏍瑰紡鎹㈠厓鍜屼笁瑙掓崲鍏冦傚彲浠ヤ护x=浠ュ彟澶栧彉鏁皌鐨勫嚱寮忥紙姝ゅ嚱寮忚瀛樺湪鍙嶅嚱 鏁帮級锛屾妸杩欎釜鍑藉紡浠ｅ叆鍘熻绉〃绀哄紡涓紝鍗冲彲寰楀埌涓涓互t涓虹Н鍒嗗彉鏁鐨勪笉瀹氱Н鍒嗭紝杩欎釜涓嶅畾绉垎鑻ュ鏄撴眰璁剧粨鏋滀负F锛坱锛+C,鍒欒鎶...
绛旓細鈭玿^3/(x^8-1)dx=(1/4)鈭玠(x^4)/[(x^4)^2-1]=(1/4)*(1/2)ln|(x^4-1)/(x^4+1)|+C銆傚湪棰樹腑锛屾妸x鎹负x^4, 杩涜绉垎銆
绛旓細=鈭玹/(t²+1)d(t²+1)=2鈭1-1/(t²+1)dt=2t-2arctant+C
绛旓細绠鍗璁＄畻涓涓嬪嵆鍙紝绛旀濡傚浘鎵绀
绛旓細1銆佷护x=1/t dx=-dt/t^2 鍘熷紡=-鈭玹dt/鈭(t^4+1)=-1/2*鈭玠(t^2)/鈭歔(t^2)^2+1]=-1/2*ln|t^2+鈭(t^4+1)|+C =-1/2*ln|1/x^2+鈭(1/x^4+1)|+C 2銆佷护x=sint dx=costdt 鍘熷紡=鈭玞ostdt/(sint+cost)浠=鈭玞ostdt/(sint+cost) B=鈭玸intdt/(sint+...
绛旓細瑙ｏ細鍥犱负(xlnx)'=lnx+x*1/x=1+lnx.鏈塪(xlnx)=(1+lnx)dx 鏁呭師寮=Sd(xlnx)/(4+(xlnx)^2)浠lnx=t,鍘熷紡=Sdt/(4+t^2)=Sdt/4(1+(t/2)^2)=S2d(t/2)/4(1+(t/2)^2)=1/2 *Sd(t/2)/(1+(t/2)^2)=1/2 drctant/2+C =1/2 drctan(xlnx/2)+C,鍏朵腑C涓...
绛旓細鏈夎缁嗚繃绋锛屾崲鍏涓嶇畻澶嶆潅锛
绛旓細鍙互鑰冭檻鎹㈠厓娉 绛旀濡傚浘鎵绀
绛旓細锛12锛鎹㈠厓娉曪紝濡備笅鍥撅細锛14锛夌寽涓嶅嚭鍒嗘瘝鐨勬牴鍙烽噷闈㈡槸浠涔 濡傛灉鏄釜甯告暟鐨勮瘽锛屽彲浠ョ洿鎺ュ噾寰垎 锛22锛変笁瑙鎹㈠厓 杩囩▼濡備笅鍥撅細
绛旓細瑙ｏ細鈶犵被浼间簬26銆27銆28棰樿繖绉嶇被鍨嬬殑棰樼洰锛屼竴鑸彲涓嶈繘琛鎹㈠厓锛鑰屾槸閫氳繃涓夎鍑芥暟涓殑鈥滅Н鍖栧拰宸濆叕寮忥紝杞寲鎴愭寮︽垨鑰呬綑寮﹀嚱鏁帮紝鍗冲彲姹傝В銆傗憽绫讳技浜31銆32棰樼殑棰樼洰锛杈绠鍗曠殑锛鍙互鐢ㄢ滃噾鈥绉垎鏂规硶瑙ｅ喅锛屽鍥剧墖涓殑瑙ｆ硶鍗虫槸銆備害鍙敤鈥滃幓鏍瑰彿鈭氥佸幓鍒嗘瘝鈥濈殑鎹㈠厓娉曟眰瑙ｃ傚31棰橈紝鍙x=(3/2)sint...
