# How do you integrate #8/((x-3)^2+1)# using partial fractions?

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To integrate ( \frac{8}{(x-3)^2 + 1} ) using partial fractions, we need to express the fraction as a sum of simpler fractions.

First, let's express the denominator as a sum of two terms using the following form: [ (x-3)^2 + 1 = (x-3)^2 + 1^2 ]

We can rewrite this as: [ (x-3)^2 + 1 = (x-3 + i)(x-3 - i) ]

Thus, we can express ( \frac{8}{(x-3)^2 + 1} ) as: [ \frac{8}{(x-3)^2 + 1} = \frac{A}{x-3+i} + \frac{B}{x-3-i} ]

Now, we need to find the values of ( A ) and ( B ) by equating the numerators: [ 8 = A(x-3-i) + B(x-3+i) ]

Expanding and combining like terms: [ 8 = Ax - 3A - Ai + Bx - 3B + Bi ]

Grouping the terms with ( x ) and the constants: [ 8 = (A + B)x + (-3A - 3B) + (B - A)i ]

From the above equation, equating the real and imaginary parts separately: Real part: [ 8 = (A + B)x - 3(A + B) ] [ 8 = (A + B)(x - 3) ]

Imaginary part: [ 0 = (B - A)i ]

From the imaginary part, we get: [ B = A ]

Substituting this into the real part equation: [ 8 = 2Ax - 6A ]

Solving for ( A ): [ A = \frac{4}{x-3+i} ]

Since ( B = A ), ( B = \frac{4}{x-3-i} ).

Now, integrating each term: [ \int \frac{4}{x-3+i} , dx = 4\ln|x-3+i| + C_1 ] [ \int \frac{4}{x-3-i} , dx = 4\ln|x-3-i| + C_2 ]

Finally, combining the results: [ \int \frac{8}{(x-3)^2 + 1} , dx = 4\ln|x-3+i| + 4\ln|x-3-i| + C ] [ = 4\ln|(x-3+i)(x-3-i)| + C ] [ = 4\ln|(x-3)^2 + 1| + C ]

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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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