How do you find the indefinite integral of #f(x) = 1/(x+1)#?
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To find the indefinite integral of ( f(x) = \frac{1}{x+1} ), you can use the technique of integration by substitution. Let ( u = x + 1 ). Then, ( du/dx = 1 ) implies ( dx = du ). Substituting these into the integral:
[ \int \frac{1}{x+1} dx = \int \frac{1}{u} du = \ln|u| + C = \ln|x + 1| + C ]
where ( C ) is the constant of integration.
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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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