# What is #f(x) = int (x^2+x)/((x+1)(x-1) ) dx# if #f(3)=2 #?

Factor the numerator of the integrand:

Separate into two fractions:

The first fraction becomes 1:

The integrals of the two terms are well know:

Solve for C:

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To find ( f(x) = \int \frac{x^2+x}{(x+1)(x-1)} , dx ) if ( f(3) = 2 ), first, compute the indefinite integral:

[ \int \frac{x^2+x}{(x+1)(x-1)} , dx = \int \left(\frac{1}{x-1} - \frac{1}{x+1}\right) , dx ]

[ = \ln|x-1| - \ln|x+1| + C ]

Given that ( f(3) = 2 ), substitute ( x = 3 ) and ( f(x) = 2 ) into the equation:

[ 2 = \ln|3-1| - \ln|3+1| + C ]

[ 2 = \ln 2 - \ln 4 + C ]

[ 2 = \ln\frac{1}{2} + C ]

[ C = 2 - \ln\frac{1}{2} ]

[ C = 2 + \ln 2 ]

So, the function ( f(x) ) is:

[ f(x) = \ln|x-1| - \ln|x+1| + 2 + \ln 2 ]

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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.

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