这几道不定积分怎么解。 这道不定积分问题求解，需要详细过程，谢谢
\u4e0d\u5b9a\u79ef\u5206\u90e8\u5206\u6c42\u6781\u9650\u7684\u51e0\u9053\u9898\u6c42\u89e3\uff0c\u6c42\u8fc7\u7a0b1\u3001\u6d1b\u6bd4\u5854\u6cd5\u5219\uff0c\u5206\u5b50 = arcsinx\uff0c\u5206\u6bcd = 2x\uff0c\u6781\u9650 = 1/2 \u3002
2\u3001\u6d1b\u6bd4\u5854\u6cd5\u5219\uff0c\u5206\u5b50 = lnx/(x+1)\uff0c\u5206\u6bcd = 2(x-1)\uff0c\u5176\u4e2d 1/2(x+1) \u6781\u9650\u4e3a 1/4\uff0c
\u7136\u540e lnx / (x+1) \u518d\u4e00\u6b21\u6c42\u5bfc\uff0c\u5206\u5b50 = 1/x\uff0c\u5206\u6bcd = 1\uff0c\u6781\u9650 = 1*1/4 = 1/4 \u3002
3\u3001\u6c42\u5bfc\uff0c\u5206\u5b50 = cos(x^2)*2x \uff0c\u5206\u6bcd = sinx+xcosx\uff0c
\u4e0a\u4e0b\u540c\u9664\u4ee5 x \u5f97 2cos(x^2) / (sinx/x + cosx) \uff0c\u6781\u9650 = 2/(1+1) = 1 \u3002
4\u3001\u6c42\u5bfc\uff0c\u5206\u5b50 = 2x*\u221a(1+x^4) \uff0c\u5206\u6bcd = 4x^3\uff0c
\u4e0a\u4e0b\u540c\u9664\u4ee5 x^3 \u5f97 2\u221a(1+1/x^4) / 4\uff0c\u6781\u9650 = 2/4 = 1/2 \u3002
=\u222bsintde^t
=sinte^t-\u222be^tdsint
=sinte^t-\u222bcostde^t
=sinte^t-coste^t+\u222be^tdcost
=sinte^t-coste^t-\u222bsintde^t
=e^t(sint-cost)/2+C
=x(sinlnx-coslnx)/2+C
第一题：上下乘以1 - cosx，
分母变为(1 + cosx)(1 - cosx) = 1 - cos²x = sin²x
然后用∫ csc²x dx = - cotx + C以及∫ cscxcotx dx = - cscx + C
第二题：先两边求导数，然后配合[lnƒ(x)]' = ƒ'(x)/ƒ(x)的形式，再两边求积分找出ƒ(x)
将e^(- x)放进d里凑成∫ ƒ(u) du的形式，然后将之前找到的ƒ(x)照样代入，就积分得结果
第三题：先找出这个积分的结果，然后代入ƒ(x)
第四题：两边求导数就出结果
第五题：添项减项法令分式变为真分数，用∫ 1/(1 + x²) dx = arctanx + C
第六题：指数化简后，直接用公式
绛旓細鍒嗘瘝鍙樹负(1 + cosx)(1 - cosx) = 1 - cos²x = sin²x 鐒跺悗鐢ㄢ埆 csc²x dx = - cotx + C浠ュ強鈭 cscxcotx dx = - cscx + C 绗簩棰橈細鍏堜袱杈规眰瀵兼暟锛岀劧鍚庨厤鍚圼lnƒ(x)]' = ƒ'(x)/ƒ(x)鐨勫舰寮忥紝鍐嶄袱杈规眰绉垎鎵惧嚭ƒ(x)灏唀^(- x)...
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绛旓細鎹㈠厓娉曡绠涓嶅畾绉垎 渚嬪鈭 鈭(x²+1) dx 浠=tanu锛屽垯鈭(x²+1)=secu锛宒x=sec²udu銆傗埆sec³udu =鈭 secudtanu =secutanu - 鈭 tan²usecudu =secutanu - 鈭 (sec²u-1)secudu =secutanu - 鈭 sec³udu + 鈭 secudu =secutanu - 鈭 ...
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绛旓細=sint+C =x/鈭(1+x^2)+C锛屽叾涓瑿鏄换鎰忓父鏁 锛2锛変护x=3sect锛屽垯dx=3secttantdt 鍘熷紡=鈭3tant/3sect*3secttantdt =鈭3tan^2tdt =3*鈭(sec^2t-1)dt =3tant-3t+C =鈭(x^2-9)-3arccos(3/x)+C锛屽叾涓瑿鏄换鎰忓父鏁 锛3锛夊師寮=鈭玡^(-x)/鈭歔e^(-2x)+1]dx =-鈭玠[e^(...
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绛旓細1.璇曡瘯瀵圭粨鏋滄眰瀵硷紝浣犺繖棰樺鏋滃鏁扮浉绛夛紝閭ｅ氨鍙互鐨勩2.娌℃湁鐢ㄥ埌涓夎褰紝sint=x,t=arcsinx, cost=cos(arcsinx).濡傛灉瑕佺敤涓夎褰細sint=x/1,灏变互t涓洪攼瑙掍綔鐩磋涓夎褰紝瀵硅竟x锛屾枩杈1,閭ｄ箞閭昏竟鈭氾紙1-x^2),杩欐椂cost=1/鈭氾紙1-x^2).3.鍚1 4.tant=x=x/1,灏变互t涓洪攼瑙掍綔鐩磋涓夎褰紝...
