# How do you integrate #1 / ((x^2 + 1) (x^2 +4))# using partial fractions?

We will use the Method of Partial Fraction to decompose the

Using Heavyside's Cover-up Method, we have,

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To integrate ( \frac{1}{(x^2 + 1)(x^2 + 4)} ) using partial fractions, we first decompose it into partial fractions of the form:

[ \frac{1}{(x^2 + 1)(x^2 + 4)} = \frac{A}{x^2 + 1} + \frac{B}{x^2 + 4} ]

where ( A ) and ( B ) are constants to be determined. Then, we solve for ( A ) and ( B ) by finding a common denominator:

[ 1 = A(x^2 + 4) + B(x^2 + 1) ]

Equating coefficients of like terms:

[ 0x^3 + 0x^2 + 1 = (A + B)x^2 + (4A + B)x + 4A ]

We equate coefficients:

[ A + B = 0 ] [ 4A + B = 0 ] [ 4A = 1 ]

Solving these equations yields:

[ A = \frac{1}{4} ] [ B = -\frac{1}{4} ]

Thus, we can write the original expression as:

[ \frac{1}{(x^2 + 1)(x^2 + 4)} = \frac{1/4}{x^2 + 1} - \frac{1/4}{x^2 + 4} ]

Now, we can integrate each term separately:

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

These integrals are standard forms and can be evaluated as follows:

[ \frac{1}{4} \int \frac{1}{x^2 + 1} , dx = \frac{1}{4} \arctan(x) + C_1 ] [ \frac{1}{4} \int \frac{1}{x^2 + 4} , dx = \frac{1}{8} \ln|x^2 + 4| + C_2 ]

Therefore, the integral of ( \frac{1}{(x^2 + 1)(x^2 + 4)} ) using partial fractions is:

[ \frac{1}{4} \arctan(x) - \frac{1}{8} \ln|x^2 + 4| + C ]

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