(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+ sin? xdsinx=(sinx)? 8/8-(sinx)? 6/3+(sinx)? 4/4+C
( 10)=∫√(4-(x+ 1)? ) dx replaces 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
(1 1) exchange? √x=u,dx=3u? Du (surname)
=∫3u? sinudu=-3∫u? dcosu=-3u? cosu+3∫cosudu? =-3u? cosu+6∫udsinu
=-3u? cosu+6usinu-6∫sinudu=-3u? cosu+6usinu+6cosu+C
(14) substitute 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