求这道题，不定积分 求这道题不定积分
\u6c42\u8fd9\u9053\u9898\u7684\u4e0d\u5b9a\u79ef\u5206
\u5982\u4e0a\u56fe\u6240\u793a\u3002
\u4ee5\u4e0a\uff0c\u8bf7\u91c7\u7eb3\u3002
则原式=∫1/[(1+2tan²u)√(1+tan²u)]dx
=∫1/[(1+2tan²u)√(1+tan²u)]dx
=∫cosu/(1+sin²u)du
=∫d(sinu)/(1+sin²u)
=arctan(sinu)+C
=arctan(sin(arctanx))+C
绛旓細(1)锛岃1/[(x-1)(x²+1)=a/(x-1)+(bx+c)銆傝В寰梐=1/2銆乥=c=-1/2銆傗埓鍘熷紡=(1/2)ln涓(x-1)+(1/4)ln(x²+1)+(1/2)arctanx+C銆(2)棰橈紝璁1-x=t锛屸埓鍘熷紡=-鈭(1-2t+t²)dt/t^100=(1/99)/t^99-(2/98)/t^98-(1/97)/t^97+C,鍏朵腑t...
绛旓細鈭 (x+1)/(x^2-2x+5) dx =(1/2)鈭 (2x-2)/(x^2-2x+5) dx +2鈭 dx/(x^2-2x+5)=(1/2)ln|x^2-2x+5| +2鈭 dx/(x^2-2x+5)=(1/2)ln|x^2-2x+5| +arctanu[( x-1)/2] + C consider x^2-2x+5 =(x-1)^2 +4 let x-1 = 2tanu dx= 2(secu)^2...
绛旓細濡備笅鍥剧墖
绛旓細鍥炵瓟锛氣埆(3x-5)³dx =1/3 鈭(3x-5)³d(3x-5) =1/12 (3x-5)^4 + C
绛旓細濡備笂鍥
绛旓細鈭氾紙1+x2) 杩欎釜褰㈠紡鐨勶紝鍙互鐢▁=tan t 鎹㈠厓鍥犱负1+tan2 t锛漵ec2t
绛旓細绗竴涓绉垎濂借薄鏂囧瓧鎺掔増鏈夐棶棰橈紝浜屽垎涔嬩簲搴旇鏄箓锛屽埄鐢╠x=d(x-2), 寰楀埌缁撴灉鏄2/7路(x-2)^(7/2)+C;绗簩涓埄鐢╠x=-1/2路d(-2x+3)锛岀粨鏋滄槸1/[2(3-2x)]+C;绗笁涓埄鐢╠x=1/2路 d(5+2x), 缁撴灉鏄2ln|5+2x|+C;鏈鍚庝竴涓埄鐢▁-1=1/2路2x-1锛屾媶鎴愪袱涓Н鍒嗙殑宸殑浜屽垎涔...
绛旓細濡傚浘 姹傞噰绾
绛旓細濡傚浘
绛旓細濡傚浘
