找一些微积分的答案！

(2)=∫4/sin？2xdx=2∫csc？2xd2x=-2cot2x+C

(3)=1/4∫(2x？+3)?(1/2)d(2x？+3)=(1/6)(2x？+3)?(3/2)+C

(4)=1/4∫sin？2xdx = 1/8∫(1-cos4x)dx = x/8-sin4x/32+C

(7)=∫sin？x(1-sin？x)？dsinx=∫(sinx)？7-2(sinx)？5+罪？xdsinx=(sinx)？8/8-(sinx)？6/3+(sinx)？4/4+C

(10)=∫√(4-(x+1)？)dx替换x=2sinu-1

=∫2co sud(2 sinu-1)= 2 ∫( cos2u+1)du = sin2u+2u+C

=(x+1)√(4-(x+1)？)/2+2arcsin((x+1)/2)+C

(11)交换？√x=u，dx=3u？杜（姓氏）

=∫3u？sinudu=-3∫u？dcosu=-3u？cosu+3∫cosudu？=-3u？cosu+6∫udsinu

=-3u？cosu+6usinu-6∫sinudu=-3u？cosu+6usinu+6cosu+C

(14)代入x=2secu，=∫2tanu/2secud2secu=2∫tan？udu=2tanu-2u=√(x？-4)-2arccos(2/x)+C

(17)=xln(x？+1)-∫xdln(x？+1)=xln(x？+1)-∫2x？/(x？+1)dx=xln(x？+1)-2x+2arctanx+C

(22)=∫x/(x-1)(x？+1)= 1/2∫(1/(x-1)-(x-1)/(x？+1))dx

= 1/2∫1/(x-1)dx-1/2∫x/(x？+1)dx+1/2∫1/(x？+1)dx

=(1/2)ln | x-1 |-(1/4)ln(x？+1)+arctanx/2+C