一道最基础的不定积分计算题
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\u7b80\u5355\u51d1\u5fae\u5206\u65b9\u6cd5
\u7528\u4e24\u6b21\u5206\u90e8\u79ef\u5206\u6cd5\uff0c\u5c31\u80fd\u7b97\u51fa\u6765\u4e86\u3002
见图
绛旓細1銆侊紳-1/3鈭(2-3x)^3/2 d(-3x)=-1/3*2/5*(2-3x)^(5/2)+C=-2/15*(2-3x)^(5/2)+C 2銆=鈭玡^(3xlna) dx=1/(3lna)*鈭玡^(3xlna) d(3xlna)=1/(3lna)*e^(3xlna)+C=1/(3lna)*a^(3x)+C 3銆=1/2鈭垰(4-x^2) d(x^2)=-1/2鈭(4-x^2)^(1/2) ...
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绛旓細鈭礳os(3x+2x)=cos3xcos2x-sin3xsin2x cos(3x-2x)=cos3xcos2x+sin3xsin2x 鈭碿os(3x+2x)+cos(3x-2x)=2cos3xcos2x 鍗砪os5x+cosx=2cos3xcos2x 鈭碿os3xcos2x=1/2*(cosx+cos5x)鈭粹埆cos3xcos2xdx =1/2*鈭(cosx+cos5x)dx =1/2*sinx+1/10*sin5x+C ...
绛旓細绠璁＄畻涓涓嬪嵆鍙紝绛旀濡傚浘鎵绀
绛旓細鍏紡锛氣埆[f(x)卤g(x)] dx = 鈭玣(x) dx 卤 鈭玤(x) dx 鈭玿^n dx = [x^(n+1)]/(n+1) + C 瀵兼暟鍙互杩樺師缁撴灉銆俤[x³/3 - (3/2)x² + 2x + C]/dx = (1/3)d(x³)/dx - (3/2)d(x²)/dx + 2d(x)/dx + d(C)/dx = (1/3)(3x&...
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绛旓細濡備笅鍥炬墍绀猴紝閮芥槸鐢ㄥ噾绉垎鐨勬柟娉曞仛
绛旓細鍥炵瓟锛氣埆f'(lnx)/xdx =鈭玣'(lnx)dlnx =f(lnx)+C =-x+C
绛旓細杩欐牱瀛愩傘傘
