高等数学，不定积分。这道题怎么解？ 高等数学，请问这道题怎么求不定积分?
\u600e\u4e48\u89e3\u8fd9\u9053\u9898\u7684\u4e0d\u5b9a\u79ef\u5206\uff1f\u539f\u5f0f=-\u222b1/\u221a(1+x²)*d(1/x)
\u4ee41/x=t,\u5219x=1/t
\u539f\u5f0f=-\u222bdt/\u221a(1+1/t²)
=-\u222btdt/\u221a(t²+1)
=-1/2*\u222b(t²+1)^(-1/2)*d(t²+1)
=-(t²+1)^(1/2)+C
=-\u221a(1+1/x²)+C
如图
根据不定积分的定义，不定积分的结果是被积函数的原函数加C，所以f(x)=(xlnx+C)'=1+lnx，答案是A。
绛旓細鏍规嵁涓嶅畾绉垎鐨勫畾涔夛紝涓嶅畾绉垎鐨勭粨鏋滄槸琚Н鍑芥暟鐨勫師鍑芥暟鍔燙锛屾墍浠(x)=(xlnx+C)'=1+lnx锛岀瓟妗堟槸A銆
绛旓細鍒嗛儴绉垎 鈭玡^xsinxdx =鈭玸inxde^x =sinx*e^x-鈭玡^xdsinx =sinx*e^x-鈭玡^xcosxdx =sinx*e^x-鈭玞osxde^x =sinx*e^x-cosx*e^x+鈭玡^xdcosx =sinx*e^x-cosx*e^x-鈭玡^xsinxdx 绉婚」 鎵浠2鈭玡^xsinxdx=sinx*e^x-cosx*e^x 鎵浠モ埆e^xsinxdx=e^x(sinx-cosx)/2 鍐嶆妸涓婁笅闄...
绛旓細=鈭(x-1)/(x^2-x+1)dx-鈭1/(x^2-x+1)dx =1/2ln(x^2-x+1)-鈭1/(x-1/2)^2+3/4)dx =1/2ln(x^2-x+1)-4/3*鈭1/(2x/鈭3-1/鈭3)^2+1)dx =1/2ln(x^2-x+1)-2/鈭3*arctan(2x/鈭3-1/鈭3)+C ...
绛旓細鏂规硶濡備笅锛岃浣滃弬鑰冿細
绛旓細鍘熷紡=鈭玿/(x²+1)dx+鈭1/(x²+1)dx =1/2路鈭1/(x²+1)路d(x²+1)+arctanx =1/2路ln(x²+1)+arctanx+C 銆愰檮娉ㄣ戔埆1/(x²+1)dx=arctanx+C 鈭玿/(x²+1)dx=1/2路ln(x²+1)+C ...
绛旓細f(X)灏辨槸閭ｄ釜寮忓瓙鐨涓嶅畾绉垎锛瑙ｅ緱 f(X)=X^3/27-2鏍瑰彿(X^3)/(3鏍瑰彿3)+C.娉ㄦ剰涓婇潰鐨勫ぇX=3x^2锛屾墍浠ュぇX鏄ぇ浜庣瓑浜庨浂鐨勩傚嵆f(x)鐨勫畾涔夊煙鍦▁>=0.鑰屽鍚堝嚱鏁癴(3x^2)鐨勫畾涔夊煙鍗村湪璐熸棤绌峰埌姝ｆ棤绌蜂笂锛屾墍浠杩欓鏈変袱绉嶆儏鍐碉紝鍙兘绛旀娌℃湁缁欎袱绉嶆儏鍐碉紝鏄濊檻娆犲懆浜嗐傚惁鍒欏氨寰楄瀹歺>0....
绛旓細1.杩欓亾楂樼瓑鏁板涓嶅畾绉垎闂姹傝В璁＄畻杩囩▼瑙佷笂鍥俱2.楂樼瓑鏁板涓嶅畾绉垎闂姹傝В璁＄畻鏃讹紝绗竴姝ワ紝灏嗗垎瀛愬垎姣嶅悓涔樹互1-sinx銆3.姝ら珮绛夋暟瀛︿笉瀹氱Н鍒嗛棶棰樻眰瑙ｇ殑绗簩姝ワ紝灏嗕笂杈圭殑绉垎鎷嗗紑鎴愪袱涓Н鍒嗭紝浠庤岀涓涓Н鍒嗙洿鎺ョ敤绉垎鍏紡锛岀浜屼釜绉垎鐢ㄥ噾寰垎锛屽嵆鎹㈠厓娉曪紝杩欐牱涓嶅畾绉垎灏辨眰鍑烘潵浜.4.杩欓噷绉垎鏃讹紝...
绛旓細瑙ｆ硶璇疯涓嬪浘锛氬湪寰Н鍒涓紝鍑芥暟鐨涓嶅畾绉垎鏄竴涓〃杈惧紡锛屽畾绉垎鏄竴涓暟銆傦紝
绛旓細鈭(0->蟺) 鈭(1-sinx) dx =鈭(0->蟺) 鈭(sin^2 x/2 +cos^2 x/2 -2sinx/2 * cosx/2) dx =鈭(0->蟺) 鈭(sinx/2 -cosx/2)^2 dx 鈭0<=x<=蟺 鈭0<=x/2<=蟺/2 褰0<=x/2<=蟺/4鏃 cosx/2>sinx/2 褰撓/4<=x/2<=蟺/2鏃 cosx/2<sinx/2 鈭 鍘熷紡=鈭(0...
绛旓細鍘熷紡=1/2*鈭2(x+1-2)dx/(x²+2x+3)=1/2*鈭(2x+2)dx/(x²+2x+3)-1/2*鈭4dx/(x²+2x+3)=1/2*鈭玠(x²+2x+3)/(x²+2x+3)-2鈭玠(x+1)/[(x+1)²+2]=1/2*ln|x²+2x+3|-鈭2*arctan[(x+1)/鈭2]+C ...
