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I have this partial fraction: $${3x+7}\over{(x-4)^2+25}$$ As far as I can tell, I do not think this can be decomposed. Is that a correct assumption?

Sorry for the very short question, there isn't much work I could show, I think.

Thank you!

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  • $\begingroup$ Yes, in $\mathbb{R}$ is this fraction irreducible. $\endgroup$
    – user376343
    Nov 8 '18 at 23:29
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In case the purpose is to find an antiderivative

$$\frac{3x+7}{x^2-8x+41} = \frac{3x-12}{x^2-8x+41} + \frac{19}{x^2-8x+41} = \left( \frac{3}{2} \right) \left( \frac{2x-8}{x^2-8x+41} \right) + \left( \frac{19}{(x-4)^2+25} \right) $$

For the 19 part, we would expect to use a substitution $$ x-4 = 5 u $$

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