How do you integrate #[(x^2 + 1)/(x^2 - 1)]dx#?
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To integrate ( \frac{x^2 + 1}{x^2 - 1} ), you can use partial fraction decomposition:
[ \frac{x^2 + 1}{x^2 - 1} = 1 + \frac{2}{x^2 - 1} ]
Then, you integrate each term separately:
[ \int \frac{x^2 + 1}{x^2 - 1} , dx = \int 1 , dx + \int \frac{2}{x^2 - 1} , dx ]
[ = x + 2 \int \frac{1}{x^2 - 1} , dx ]
For the integral ( \int \frac{1}{x^2 - 1} , dx ), you can use the substitution ( u = x^2 - 1 ):
[ du = 2x , dx ] [ dx = \frac{1}{2x} , du ]
Substituting ( u ) and ( dx ), you get:
[ \int \frac{1}{x^2 - 1} , dx = \int \frac{1}{u} \left( \frac{1}{2x} , du \right) ]
[ = \frac{1}{2} \int \frac{1}{u} , du ]
[ = \frac{1}{2} \ln|u| + C ]
[ = \frac{1}{2} \ln|x^2 - 1| + C ]
Substituting back into the original integral:
[ \int \frac{x^2 + 1}{x^2 - 1} , dx = x + \ln|x^2 - 1| + C ]
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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.
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.
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.
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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