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∫(x/(x+1)2)dx

=∫((x+1-1)/(x+1)2)dx

=∫(1/(x+1))dx - ∫(1/(x+1)2)dx

u=x+1, du=dx

∫(1/u)du - ∫(1/u2)du

=log(u) - (-1/u) + C

=log(x+1) + 1/(x+1) + C

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Q: Find the indefinite integral x divided by x plus 1 quantity squared dx?
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