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Find some answers to calculus!
( 1)=2∫cos√xd√x=2sin√x+C

(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