不定积分1咋求
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绛旓細绗竴绉嶈В娉曚腑锛岀孩鑹叉鍐呭彉鎹㈤敊璇紝搴旇涓猴細缁胯壊鍦堝唴閮ㄥ垎锛屽彲浠ヨ繘琛屽彉鎹細鍥犳鏈夛細涓ょ鏂规硶缁撴灉涓鑷达紝浣犲ぇ鎰忎簡銆
绛旓細鏂规硶锛氱啛缁冪殑璇濆彲浠ョ洿鎺ョ敤x=1-(1-x)灞曞紑鍚绉垎锛屽綋鐒讹紝t=鈭1-x 鎹㈠厓涔熸槸甯歌瑙ｆ硶 杩囩▼锛氬弬鑰冧笅鍥
绛旓細鍒╃敤鍑戝井鍒嗘硶鍙互鐩存帴姹傚嚭缁撴灉銆
绛旓細瑙ｇ瓟杩囩▼濡備笅:
绛旓細鍋囪鏄1/sinx鐨绉垎 鈭1/sinx dx =鈭1/[2sin(x/2)cos(x/2)] dx,涓ゅ嶈鍏紡 =鈭1/[sin(x/2)cos(x/2)] d(x/2)=鈭1/tan(x/2)*sec²(x/2) d(x/2)=鈭1/tan(x/2) d[tan(x/2)], [娉ㄢ埆sec²(x/2)d(x/2)=tan(x/2)+C]=ln|tan(x/2)|+C, ...
绛旓細鈭 dx/[x(1+x⁴)]浠=x⁴,du=4x³ dx 鍘熷紡= 鈭 1/[x*(1+u)] * du/(4x³)= (1/4)鈭 1/[u(u+1)] du = (1/4)鈭 (u+1-u)/[u(u+1)] du = (1/4)鈭 [1/u - 1/(u+1)] du = (1/4)(ln|u| - ln|u+1|) + C = (1/4)...
绛旓細鈭玔-1,1] x^11 cosxdx =鈭玔-1,0]x^11cosxdx+鈭玔0,1]x^11cosxdx =-鈭玔0,-1]x^11cosxdx+鈭玔0,1]x^11cosxdx 鈭玔0,-1]x^11cosxdx ( x=-u, x=-1,u=1) =鈭玔0,1] -u^11cosud(-u)=鈭玔0,1}u^11cosudu =-鈭玔0,1}u^11cosudu+鈭玔0,1]x^11cosxdx...
绛旓細浠(x)=e^x*tanx/2锛屸啋涓婂紡=鈭玠f(x)锛宒f(x)=f'(x)dx锛屸啋涓婂紡=鈭玠f(x)=鈭玣'(x)dx=f(x)+c
绛旓細瑙ｆ瀽濡備笅锛氣埆 1/sinx dx = 鈭 cscx dx = 鈭 cscx * (cscx - cotx)/(cscx - cotx) dx = 鈭 (- cscxcotx + csc²x)/(cscx - cotx) dx = 鈭 d(cscx - cotx)/(cscx - cotx)= ln|cscx - cotx| + C 璁綟(x)鏄嚱鏁癴(x)鐨涓涓鍘熷嚱鏁锛屽嚱鏁癴(x)鐨勬墍鏈夊師鍑芥暟F(x)+ ...
