不定积分题求解
\u4e0d\u5b9a\u79ef\u5206\u7684\u9898\u6c42\u89e3\u3000\u3000\u89e3\uff1a2\u9898\uff0c\u8bbex=tan\u03b8\uff0c\u2234\u539f\u5f0f=\u222bcos\u03b8d\u03b8=sin\u03b8+C=x(1+x^2)^(-1/2)+C\u3002
\u3000\u30003\u9898\uff0c\u8bbex=sec\u03b8\uff0c\u2234\u539f\u5f0f=\u222bcos\u03b8d\u03b8=sin\u03b8+C=(1/x)(x^2-1)^(1/2)+C\u3002
\u3000\u30001\u9898\uff0c\u8bbex=(1/3)tan\u03b8\uff0c\u2234\u539f\u5f0f=(1/3)\u222bsec\u03b8d\u03b8=(1/3)ln\u4e28sec\u03b8+tan\u03b8\u4e28+C=(1/3)ln\u4e28(9x^2+1)^(1/2)+3x\u4e28+C\u3002\u2234\u9009A\u3002
\u3000\u3000\u4f9b\u53c2\u8003\u3002
\u5982\u56fe
∫ln(2x－3) dx
＝xln(2x－3)－∫2x/(2x－3) dx
＝xln(2x－3)－∫[1＋3/(2x－3)]dx
＝xln(2x－3)－x－3/2×∫1/(2x－3)d(2x－3)
＝xln(2x－3)－x－3/2×ln(2x－3)＋C
2、∫2sin(2x－3)dx＝∫sin(2x－3)d(2x－3)＝－cos(2x－3)＋C
3、∫1/(3－2x)dx＝－1/2×∫1/(3－2x)d(3－2x)＝－1/2×ln|3－2x|＋C
4、∫(3－2x)^3 dx＝－1/2×∫(3－2x)^3 d(3－2x)＝－1/2×1/4×(3－2x)^4＋C＝－1/8×(3－2x)^4＋C
5、因为∫tanx dx＝－ln|cosx|＋C，所以，
∫tan(2x－3)dx＝1/2×∫tan(2x－3)d(2x－3)＝－1/2×ln|cos(2x－3)|＋C
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绛旓細鍘熷紡=x(arcsinx)^2-鈭玿d[(arcsinx)^2]=x(arcsinx)^2-鈭2xarcsinx/鈭(1-x^2)dx =x(arcsinx)^2+鈭2arcsinxd[鈭(1-x^2)]=x(arcsinx)^2+2鈭(1-x^2)arcsinx-鈭2鈭(1-x^2)d(arcsinx)=x(arcsinx)^2+2鈭(1-x^2)arcsinx-鈭2dx =x(arcsinx)^2+2鈭(1-x^2)arcsinx...
绛旓細鈭玿f'(x) dx = ln(1+x^2) +c 涓よ竟姹傚 xf'(x) = 2x/(1+x^2)f'(x) = 2/(1+x^2)f(x) =鈭 2/(1+x^2) dx = 2arctanx + C'f(1)=2 2=2(蟺/4) +C'C'= 2-蟺/2 f(x) = 2arctanx +2 -蟺/2 ...
绛旓細1 鈭玜rcsinxdx/x^2 =鈭玜rcsinxd(-1/x)= -arcsinx/x+鈭玠x/x鈭(1-x^2)x=sinu 鈭玠x/x鈭(1-x^2)=鈭玠u/sinu=鈭-dcosu/(1-cosu)(1+cosu)=(-1/2)ln|(1+cosu)/(1-cosu)| =ln|(1-cosu)/sinu|+C =ln|1/x-鈭(1-x^2)/x|+C =-arcsinx/x+ln|1/x-鈭(1-x^2)/...
绛旓細棣栧厛闄嶅箓 =x(1-cos2x)/2 鍏朵腑锛寈cos2xdx=0.5xd(sin2x)=0.5xsin2x+0.5sin2xdx锛绉垎鍗冲彲
绛旓細鍏堝cos²x杩涜闄嶆锛屽埄鐢ㄤ簩鍊嶈鍏紡cos2x=2cos²x-1杩涜闄嶆锛岀劧鍚庡啀鍒嗘垚涓ら」绉垎鐨勫拰锛屽墠闈㈢殑绠鍗曪紝鍚庨潰鐨勬妸涓夎鍑芥暟鍑戝井鍒嗭紝涓ゆ鍒嗛儴绉垎瀵箈²闄嶆锛屾渶鍚庡緱鍒扮瓟妗堛
绛旓細涓嶅畾绉垎棰,姹傝В,璋㈣阿銆 鎴戞潵绛 1涓洖绛 #鐑# 楣ゅ矖鐖嗙伀鑳屽悗鐨勫師鍥犳槸浠涔?瀣村偛71 2018-11-30 路 TA鑾峰緱瓒呰繃114涓禐 鐭ラ亾绛斾富 鍥炵瓟閲:148 閲囩撼鐜:100% 甯姪鐨勪汉:40涓 鎴戜篃鍘荤瓟棰樿闂釜浜洪〉 鍏虫敞 灞曞紑鍏ㄩ儴 鈭玔1/(a²sin²x+b²cos²x)]dx=鈭玔(1/cos²x)/(a²sin...
绛旓細浠も垰x=t x=t骞虫柟 dx=2tdt 鎵浠 鍘熷紡=2鈭玹e^tdt =2鈭玹de^t =2te^t-2鈭玡^tdt =2te^t-2e^t+c =2鈭歺路e^(鈭歺)-2e^(鈭歺)+c
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