求几道微积分的答案！

Question

求几道微积分的答案！大佬学霸们拜托了( •̥́ ˍ •̀ू )

Answer 1

(1)=2∫cos√xd√x=2sin√x+C
(2)=∫4/sin²2xdx=2∫csc²2xd2x=-2cot2x+C
(3)=1/4∫(2x²+3)ˆ(1/2)d(2x²+3)=(1/6)(2x²+3)ˆ(3/2)+C
(4)=1/4∫sin²2xdx=1/8∫(1-cos4x)dx=x/8-sin4x/32+C
(7)=∫sin³x(1-sin²x)²dsinx=∫(sinx)ˆ7-2(sinx)ˆ5+sin³xdsinx=(sinx)ˆ8/8-(sinx)ˆ6/3+(sinx)ˆ4/4+C
(10)=∫√(4-(x+1)²)dx换元x=2sinu-1
=∫2cosud(2sinu-1)=2∫(cos2u+1)du=sin2u+2u+C
=(x+1)√(4-(x+1)²)/2+2arcsin((x+1)/2)+C
(11)换元³√x=u，dx=3u²du
=∫3u²sinudu=-3∫u²dcosu=-3u²cosu+3∫cosudu²=-3u²cosu+6∫udsinu
=-3u²cosu+6usinu-6∫sinudu=-3u²cosu+6usinu+6cosu+C
(14)换元x=2secu，=∫2tanu/2secud2secu=2∫tan²udu=2tanu-2u=√(x²-4)-2arccos(2/x)+C
(17)=xln(x²+1)-∫xdln(x²+1)=xln(x²+1)-∫2x²/(x²+1)dx=xln(x²+1)-2x+2arctanx+C
(22)=∫x/(x-1)(x²+1)=1/2∫(1/(x-1)-(x-1)/(x²+1))dx
=1/2∫1/(x-1)dx-1/2∫x/(x²+1)dx+1/2∫1/(x²+1)dx
=(1/2)ln|x-1|-(1/4)ln(x²+1)+arctanx/2+C追问

稳了，大佬