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How do you find #int (x-1)/sqrt(x^2-2x)#?

Answer 1

Aurora Alvarado

#sqrt(x^2-2x)+C#

#I=int(x-1)/sqrt(x^2-2x)dx=1/2int(2x-2)/sqrt(x^2-2x)dx#

Substituting #u=x^2-2x# yields #du=(2x-2)dx#.

#I=1/2int(du)/sqrtu=1/2intu^(-1/2)du#

With the rule #intu^ndu=u^(n+1)/(n+1)+C#, we see that

#I=1/2u^(1/2)/(1/2)+C=u^(1/2)+C=sqrtu+C=sqrt(x^2-2x)+C#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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