# How do you use partial fractions to find the integral #int (x^2+x+3)/(x^4+6x^2+9)dx#?

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Noticing that the denominator is a perfect square, we see there is no need to use partial fractions.

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First, factor the denominator: ( x^4 + 6x^2 + 9 = (x^2 + 3)^2 ). Then, express the integrand as partial fractions:

[ \frac{x^2 + x + 3}{x^4 + 6x^2 + 9} = \frac{Ax + B}{x^2 + 3} + \frac{Cx + D}{(x^2 + 3)^2} ]

Multiply both sides by the denominator ( x^4 + 6x^2 + 9 = (x^2 + 3)^2 ) to clear the fractions:

[ x^2 + x + 3 = (Ax + B)(x^2 + 3) + (Cx + D) ]

Expand and collect like terms:

[ x^2 + x + 3 = Ax^3 + 3Ax + Bx^2 + 3B + Cx + D ]

Now, equate coefficients for corresponding powers of ( x ):

For ( x^3 ): ( A = 0 )

For ( x^2 ): ( B = 1 )

For ( x ): ( 3A + C = 1 )

For the constant term: ( 3B + D = 3 )

Solving these equations gives ( A = 0, B = 1, C = -3, D = 6 ). Now, rewrite the integrand using these partial fractions:

[ \frac{x^2 + x + 3}{x^4 + 6x^2 + 9} = \frac{x}{x^2 + 3} - \frac{3x + 6}{(x^2 + 3)^2} ]

Now integrate term by term:

[ \int \frac{x}{x^2 + 3} , dx - \int \frac{3x + 6}{(x^2 + 3)^2} , dx ]

For the first integral, use a substitution ( u = x^2 + 3 ) to get ( \frac{1}{2} \ln|x^2 + 3| + C_1 ).

For the second integral, use a substitution ( u = x^2 + 3 ) and apply integration by parts to get ( -\frac{1}{2(x^2 + 3)} - \frac{3}{2} \ln|x^2 + 3| + C_2 ).

Combine the results to get the final integral:

[ \frac{1}{2} \ln|x^2 + 3| - \frac{1}{2(x^2 + 3)} - \frac{3}{2} \ln|x^2 + 3| + C ]

Simplify to obtain the answer:

[ \frac{1}{2} \ln|x^2 + 3| - \frac{1}{2(x^2 + 3)} - \frac{3}{2} \ln|x^2 + 3| + C = -\frac{1}{2} \ln|x^2 + 3| - \frac{1}{2(x^2 + 3)} + C ]

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