# How do you integrate 1/ (x^2 -2x + 17)?

Apr 12, 2015

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

Method
The idea behind this one is: you want to make that base( denominator: ${x}^{2} - 2 x + 17$ ) look something like $\frac{1}{{x}^{2} + 1}$
because the integral(primitive) of that is ${\tan}^{-} 1 x$ !

So this means that we can use the substitution of $\tan$

You'll surely get that intuition as we move along!

Solution

${x}^{2} - 2 x + 17 = {\left(x - 1\right)}^{2} - 1 + 17$ By completing the square

$= {\left(x - 1\right)}^{2} + 16$

$= 16 \left[\frac{{\left(x - 1\right)}^{2}}{16} + 1\right] = 16 \left[{\left(\frac{x - 1}{4}\right)}^{2} + 1\right]$

$\implies \frac{1}{{x}^{2} - 2 x + 17} = \frac{1}{16 \left[{\left(\frac{x - 1}{4}\right)}^{2} + 1\right]}$

Therefore,

$\int \frac{1}{{x}^{2} - 2 x + 17} \text{dx" = 1/16int 1/(((x - 1)/4)^2 + 1)"dx}$

<Now that looks more like $\int \frac{1}{{x}^{2} + 1}$>

So we let $\frac{x - 1}{4} = \tan \theta$

$\implies \frac{\mathrm{dx}}{4} = {\sec}^{2} \theta d \theta \implies \mathrm{dx} = 4 {\sec}^{2} \theta d \theta$

$\implies \frac{1}{16} \int \frac{1}{{\tan}^{2} \theta + 1} \cdot 4 {\sec}^{2} \theta d \theta$

$\implies \frac{4}{16} \int \frac{\cancel{{\sec}^{2} \theta}}{\cancel{{\sec}^{2} \theta}} d \theta$

$\implies \frac{1}{4} \int d \theta = \frac{1}{4} \theta + C$

Remember we let, $\frac{x - 1}{4} = \tan \theta \implies \theta = {\tan}^{-} 1 \left(\frac{x - 1}{4}\right)$

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