# How do you integrate 8/(x^2-8x+17) using partial fractions?

Jul 27, 2016

$8 {\tan}^{- 1} \left(x - 4\right) + C$

#### Explanation:

Since ${x}^{2} - 8 x + 17 = {x}^{2} - 2 \times 4 x + {4}^{2} + 1 = {\left(x - 4\right)}^{2} + 1$ has no real zeros, it can not be factorized without using complex numbers. This is not a problems as far as evaluating the integral is concerned, since we can use the substitution $t = x - 4$ makes its evaluation rather simple:

$\int \frac{8}{{x}^{2} - 8 x + 17} \mathrm{dx} = \int \frac{8}{{\left(x - 4\right)}^{2} + 1} \mathrm{dx} = \int \frac{8}{1 + {t}^{2}} \mathrm{dt} = 8 {\tan}^{- 1} t + C = 8 {\tan}^{- 1} \left(x - 4\right) + C$

If we insist on using partial fractions to reduce the integral further ( though I wouldn't particularly recommend it ), we can make use of

${x}^{2} - 8 x + 17 = \left(x - 4 + i\right) \left(x - 4 - i\right)$

so that

$\frac{1}{x - 4 - i} - \frac{1}{x - 4 + i} = \frac{2 i}{{x}^{2} - 8 x + 17}$

and thus

$\int \frac{8}{{x}^{2} - 8 x + 17} \mathrm{dx} = \frac{4}{i} \int \left(\frac{1}{x - 4 - i} - \frac{1}{x - 4 + i}\right) \mathrm{dx} = \frac{4}{i} \left(\ln \left(x - 4 - i\right) - \ln \left(x - 4 + i\right)\right) + C ' = \frac{4}{i} \ln \left(\frac{x - 4 - i}{x - 4 + i}\right) + C ' = \frac{4}{i} \ln \left(\frac{- i}{i} \frac{1 + i \left(x - 4\right)}{1 - i \left(x - 4\right)}\right) + C ' = \frac{4}{i} \ln \left(\frac{1 + i \left(x - 4\right)}{1 - i \left(x - 4\right)}\right) + C ' + \frac{4}{i} \ln \left(- 1\right)$

To simplify this further we have to resort to the polar form :

$1 \pm i \left(x - 4\right) = \sqrt{{\left(x - 4\right)}^{2} + 1} \exp \left(\pm i {\tan}^{- 1} \left(x - 4\right)\right)$

so that

$\frac{1 + i \left(x - 4\right)}{1 - i \left(x - 4\right)} = \exp \left(2 i {\tan}^{- 1} \left(x - 4\right)\right)$

and the integral becomes

$\frac{4}{i} 2 i {\tan}^{- 1} \left(x - 4\right) + C$

where $C = C ' + 4 \pi$