How do you integrate int 1/sqrt(x^2-4x+13)dx using trigonometric substitution?

Mar 10, 2016

$\int \frac{1}{\sqrt{{x}^{2} - 4 x + 13}} = l n | \sqrt{1 + {\left(x - 2\right)}^{2} / 9} + \frac{x - 2}{3} | + C$

Explanation:

$\int \frac{1}{\sqrt{{x}^{2} - 4 x + 13}} d x = \int \frac{1}{\sqrt{{x}^{2} - 4 x + 9 + 4}} d x$
$\int \frac{1}{\sqrt{{\left(x - 2\right)}^{2} + {3}^{2}}} d x$
$x - 2 = 3 \tan \theta \text{ } d x = 3 {\sec}^{2} \theta d \theta$
$\int \frac{1}{\sqrt{{x}^{2} - 4 x + 13}} d x = \int \frac{3 {\sec}^{2} \theta d \theta}{\sqrt{9 {\tan}^{2} \theta + 9}} = \int \frac{3 {\sec}^{2} \theta d \theta}{3 \sqrt{1 + {\tan}^{2} \theta}} \text{ } 1 + {\tan}^{2} \theta = {\sec}^{2} \theta$
$\int \frac{1}{\sqrt{{x}^{2} - 4 x + 13}} d x = \int \frac{3 {\sec}^{2} \theta d \theta}{3 \sqrt{{\sec}^{2} \theta}}$
$\int \frac{1}{\sqrt{{x}^{2} - 4 x + 13}} d x = \int \frac{\cancel{3 {\sec}^{2} \theta} d \theta}{\cancel{3 \sec \theta}}$
$\int \frac{1}{\sqrt{{x}^{2} - 4 x + 13}} d x = \int \sec \theta d \theta$
$\int \frac{1}{\sqrt{{x}^{2} - 4 x + 13}} d x = l n | \sec \theta + \tan \theta | + C$
$\tan \theta = \frac{x - 2}{3} \text{ } \sec \theta = \sqrt{1 + {\tan}^{2} \theta} = \sqrt{1 + {\left(x - 2\right)}^{2} / 9}$
$\int \frac{1}{\sqrt{{x}^{2} - 4 x + 13}} = l n | \sqrt{1 + {\left(x - 2\right)}^{2} / 9} + \frac{x - 2}{3} | + C$

Mar 10, 2016

${\sinh}^{-} 1 \left(\frac{x - 2}{3}\right) + C$

Explanation:

The hyperbolic version is also possible:

• $x - 2 = 3 \sinh u$
• $\mathrm{dx} = 3 \cosh u \mathrm{du}$

$\int \frac{1}{\sqrt{{x}^{2} - 4 x + 13}} \mathrm{dx} = \int \frac{1}{\sqrt{9 {\sinh}^{2} u + 9}} 3 \cosh u \mathrm{du} = \int \frac{1}{3 \cosh u} 3 \cosh u \mathrm{du} = u + C$

Hence:

$\int \frac{1}{\sqrt{{x}^{2} - 4 x + 13}} \mathrm{dx} = {\sinh}^{-} 1 \left(\frac{x - 2}{3}\right) + C$