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

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To find the function ( f(x) ), we need to evaluate the indefinite integral of ( \frac{{x^2 + x}}{{(x+3)(x-1)}} ) with respect to ( x ). Then we use the given information ( f(-2) = 2 ) to determine the constant of integration.

First, let's find the indefinite integral:

[ \int \frac{{x^2 + x}}{{(x+3)(x-1)}} , dx ]

To integrate this, we can use partial fraction decomposition. We express ( \frac{{x^2 + x}}{{(x+3)(x-1)}} ) as the sum of two simpler fractions:

[ \frac{{x^2 + x}}{{(x+3)(x-1)}} = \frac{A}{x+3} + \frac{B}{x-1} ]

Then, we solve for ( A ) and ( B ) by equating coefficients:

[ x^2 + x = A(x-1) + B(x+3) ]

By expanding and comparing coefficients, we find ( A = -\frac{1}{4} ) and ( B = \frac{5}{4} ). So,

[ \int \frac{{x^2 + x}}{{(x+3)(x-1)}} , dx = \int \left( -\frac{1}{4(x+3)} + \frac{5}{4(x-1)} \right) , dx ]

[ = -\frac{1}{4} \ln|x+3| + \frac{5}{4} \ln|x-1| + C ]

Now, we'll use the given condition ( f(-2) = 2 ) to determine the constant ( C ):

[ f(-2) = -\frac{1}{4} \ln|-2+3| + \frac{5}{4} \ln|-2-1| + C = 2 ]

[ \frac{1}{4} \ln 1 - \frac{5}{4} \ln 3 + C = 2 ]

[ C = 2 + \frac{1}{4} \ln 1 - \frac{5}{4} \ln 3 ]

[ C = 2 - \frac{5}{4} \ln 3 ]

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

[ f(x) = -\frac{1}{4} \ln|x+3| + \frac{5}{4} \ln|x-1| + 2 - \frac{5}{4} \ln 3 ]

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