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Question Number 483 by 13/NaSaNa(N)056565 last updated on 25/Jan/15
∫1x−1dx
Answered by 123456 last updated on 12/Jan/15
∫dxx−1u=x−1du=dx∫duu=∫u−1/2du=11/2u1/2+c=2u+c=2x−1+c
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