高等数学求不定积分，有解析就加悬赏！ 高等数学，不定积分。 有三问，会几个就解析几个吧，谢谢。
\u9ad8\u7b49\u6570\u5b661 \u6c42\u4e0d\u5b9a\u79ef\u5206 \u60ac\u8d4f\u9ad8\u5206\u56fe\u7247
(1) \u539f\u5f0f=1/2 \u222bsin²2x dsin2x
=1/2 \u00d71/(2+1) (sin2x)^(2+1)+c
=1/6 sin³2x+c
(2)\u4f7f\u7528\u7684\u662f\uff1a
\u222b1/(x²-a²) dx=1/2a \u00b7ln|(x-a)/(x+a)|+c \u516c\u5f0f
\uff083\uff09\u5229\u7528\u7684\u662f\u79ef\u5316\u548c\u5dee\u516c\u5f0f\uff1a
cosacosb=1/2 [cos(a+b)+cos(a-b)]
令x+3/2=(5/2)*sect，dx=(5/2)*secttantdt
原式=∫(5/2)*secttantdt/{[(5/2)*sect-(3/2)]*(5/2)*tant}
=∫2sectdt/(5sect-3)
=∫2dt/(5-3cost)
令u=tan(t/2)，则cost=(1-u^2)/(1+u^2)，dt=2du/(1+u^2)
原式=∫[4du/(1+u^2)]/[5-3(1-u^2)/(1+u^2)]
=∫4du/(5+5u^2-3+3u^2)
=∫2du/(1+4u^2)
=arctan(2u)+C
=arctan[2arctan(t/2)]+C
=t/[1-(t/2)^2]+C
=4t/(4-t^2)+C
={4arccos[5/(2x+3)]}/{4-[arccos[5/(2x+3)]]^2}+C，C是任意常数
绛旓細鍘熷紡=鈭玠x/x鈭歔(x+3/2)^2-25/4]浠+3/2=(5/2)*sect锛宒x=(5/2)*secttantdt 鍘熷紡=鈭(5/2)*secttantdt/{[(5/2)*sect-(3/2)]*(5/2)*tant} =鈭2sectdt/(5sect-3)=鈭2dt/(5-3cost)浠=tan(t/2)锛屽垯cost=(1-u^2)/(1+u^2)锛宒t=2du/(1+u^2)鍘熷紡=鈭玔4...
绛旓細鈭玿²e^(-2x)dx =(-½)鈭玿²d[e^(-2x)]=(-½)x²路e^(-2x) +½鈭玔e^(-2x)]d(x²)=(-½)x²路e^(-2x) +½鈭玔2x路e^(-2x)]dx =(-½)x²路e^(-2x) +鈭玔x路e^(-2x)]dx =(-½)x...
绛旓細鈭玔x³/(x+3)]dx涓嶅畾绉垎瑙ｇ瓟濡備笅锛氭涓烘湁鐞嗗垎寮忕殑绉垎銆
绛旓細璇︾粏杩囩▼濡傚浘鎵绀衡︹︹
绛旓細2. 姹備笉瀹氱Н鍒鍏紡锛- 鈭0*dx = c - 鈭玿^u*dx = (x^(u+1))/(u+1) + c - 鈭1/x*dx = ln|x| + c - 鈭玜^x*dx = (a^x) / ln(a) + c - 鈭玡^x*dx = e^x + c - 鈭玸in(x)*dx = -cos(x) + c - 鈭玞os(x)*dx = sin(x) + c - 鈭1/(cos(x))...
绛旓細鍥炵瓟锛氭牴鎹鐩湁鈭玣(x)dx= ln x/x 涓よ竟瀵箈姹傚鏈 f(x)=(1-ln x)/x² 鍘熷紡=鈭玿 d f(x) =x f(x) - 鈭玣(x) dx =(1-ln x)/x -ln x/x =(1-2ln x)/x
绛旓細(1) 鍘熷紡=1/2 鈭玸in²2x dsin2x =1/2 脳1/(2+1) (sin2x)^(2+1)+c =1/6 sin³2x+c (2)浣跨敤鐨勬槸锛氣埆1/(x²-a²) dx=1/2a 路ln|(x-a)/(x+a)|+c 鍏紡 锛3锛夊埄鐢ㄧ殑鏄Н鍖栧拰宸叕寮忥細cosacosb=1/2 [cos(a+b)+cos(a-b)]
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绛旓細鍩烘湰绉垎琛ㄥ彲鏌 (1)鈭玠x/x鈭氾紙x^2-a^2锛=1/a arccos(a/x)+C (2)鈭玠x/锛堚垰锛坅^2-x^2锛塣3锛=-x/[a^2鈭氾紙a^2-x^2锛塢+C (3)鈭玠x/鈭(a+bx+cx^2)=1/鈭歝 *ln|2cx+b+2*鈭歝*(a+bx+cx^2)|+C (c>0)(1)鈭玠x/x鈭(x^2-1)=鈭玠x/x^2鈭(1-1/x^2)=鈭...
