求不定积分。谢谢 求不定积分谢谢
\u6c42\u4e0d\u5b9a\u79ef\u5206\uff0c\u5177\u4f53\u6b65\u9aa4\uff0c\u8c22\u8c22\uff01
\u5982\u56fe \u91c7\u7eb3\u4e00\u4e0b\u8c22\u8c22
先凑微分，然后再用分部积分法就可以了
原式＝1/2 ∫ ln(1＋x²) d(1＋x²)
＝1/2 (1＋x²)ln(1＋x²) - ∫xdx
＝1/2 (1＋x²)ln(1＋x²) - 1/2 x²＋C
绛旓細浠 鈭歔(x+1)/x] = u, 鍒 x+1 = xu^2, x = 1/(u^2-1), dx = -2udu/(u^2-1)^2,I = 鈭 (u^2-1)u(-2u)du/(u^2-1)^2 = -2鈭 u^2 du/(u^2-1) = -2鈭 [1+1/(u^2-1)]du = -2u + 鈭 [1/(u-1)-1/(u+1)]du = -2u + ln|(u-1)...
绛旓細瑙ｇ瓟濡備笅:
绛旓細1,璁総=arctanx,鍖栫畝寰:鈭玸int*e^t*dt,鍦ㄨ繛缁2娆″垎甯绉垎鐨勭粨鏋滀负[(sint+cost)/2]*e^t+C =(1/2)*(1+x^2)^(1/2)*e^(arctanx)+C 2,璁锯垰[锛1-鈭歺锛/锛1+鈭歺锛塢=t锛屽寲绠鐨勶紙-8锛夆埆t^2(1-t^2)dt/(1+t^2)^3,鍦ㄦ崲鍏冿紝璁総=tanu,鍖栫畝鐨勶細鈭(sinu)^4du=鈭玔3/8...
绛旓細(1)let u=e^x du= e^x dx 鈭 xe^x/(1+e^x)^2 dx =鈭 lnu /(1+u)^2 du =-鈭 lnu d[1/(1+u)]=-lnu/(1+u) + 鈭 du/[u(1+u)]=-lnu/(1+u) + 鈭 [1/u-1/(1+u)] du =-lnu/(1+u) + ln|u| - ln|1+u| + C =-x/(1+e^x) + x - ln|1+...
绛旓細鎹㈠厓娉曪細璁3x=2y, y=(3/2)x, x=(2/3)y 鈭垰(1+sin3x锛 dx=鈭(2/3)鈭(1+sin2y锛塪y 1+sin2y=sin^2(y)+cos^2(y)+2siny*cosy=(siny+cosy)^2 鈭(2/3)鈭(1+sin2y锛塪y=鈭(2/3)(siny+cosy)dy=(2/3)(siny-cosy)+D ...
绛旓細姹傞噰绾筹綖鈭1/锛坰inx+cosx锛塪x =(鈭2/2)鈭1/[(鈭2/2)sinx+(鈭2/2)cosx] dx =(鈭2/2)鈭1/sin(x+蟺/4) dx =(鈭2/2)鈭玞sc(x+蟺/4) dx =(鈭2/2)ln|csc(x+蟺/4)-cot(x+蟺/4)|+C
绛旓細浠も垰2x-1+1=t 鍒檟=[(t-1)^2+1]/2 dx=t-1 鍒欏師寮= 鈭1/t*锛坱-1)dt =鈭(1-1/t)dt =t-lnt+C 鍙嶅甫鍥炲幓 =[(t-1)^2+1]/2+ln[(t-1)^2+1]/2+C
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绛旓細鐢ㄤ笁瑙掍唬鎹㈠彲浠ユ眰鍑虹粨鏋溿
绛旓細濡傚浘
