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答:
原式
=∫[(x+2)/[2(x+1)^2]-x/[2(x^2+1)] dx
=1/2*∫[(x+1+1)/(x+1)^2-x/(x^2+1) dx
=1/2*∫(1/(x+1)+1/(x+1)^2-x/(x^2+1))dx
=1/2*[ln|x+1|-1/(x+1)-1/2*ln(x^2+1)]+C
=1/2*ln|(x+1)/√(x^2+1)|-1/2(x+1)+C
原式
=∫[(x+2)/[2(x+1)^2]-x/[2(x^2+1)] dx
=1/2*∫[(x+1+1)/(x+1)^2-x/(x^2+1) dx
=1/2*∫(1/(x+1)+1/(x+1)^2-x/(x^2+1))dx
=1/2*[ln|x+1|-1/(x+1)-1/2*ln(x^2+1)]+C
=1/2*ln|(x+1)/√(x^2+1)|-1/2(x+1)+C
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