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(1)=∫1/(x+1)-2/(x+1)²dx
=ln|x+1|+2/(x+1)+C
(2)=∫2/(x-2)+3/(x+5)dx
=2ln|x-2|+3ln|x+5|+C
(3)=∫1/x-x/(x²+1)dx
=ln|x|-(1/2)ln(x²+1)+C
(4)=∫1/x+2/(x-1)²dx
=ln|x|-2/(x-1)+C
(5)=∫1/x-3/2(x+1)-(x+1)/2(x²+1)dx
=ln|x|-(3/2)ln|x+1|-(1/4)ln(x²+1)-(1/2)arctanx+C
=ln|x+1|+2/(x+1)+C
(2)=∫2/(x-2)+3/(x+5)dx
=2ln|x-2|+3ln|x+5|+C
(3)=∫1/x-x/(x²+1)dx
=ln|x|-(1/2)ln(x²+1)+C
(4)=∫1/x+2/(x-1)²dx
=ln|x|-2/(x-1)+C
(5)=∫1/x-3/2(x+1)-(x+1)/2(x²+1)dx
=ln|x|-(3/2)ln|x+1|-(1/4)ln(x²+1)-(1/2)arctanx+C
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