# How to integrate 1/sqrt(x^2 - 4x + 3) dx ?

Mar 27, 2018

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - 4 x + 3}} = \ln \left\mid x - 2 + \sqrt{{x}^{2} - 4 x + 3} \right\mid + C$

#### Explanation:

Complete the square at the denominator:

${x}^{2} - 4 x + 3 = {\left(x - 2\right)}^{2} - 1$

and subtitute:

$x - 2 = \sec t$

$\mathrm{dx} = \tan t \sec t \mathrm{dt}$

Note also that:

${x}^{2} - 4 x + 3 = \left(x - 3\right) \left(x - 1\right)$

and the integrand function is defined for

${x}^{2} - 4 x + 3 > 0$

that is for $x \in \left(- \infty , 1\right) \cup \left(3 , + \infty\right)$. We will start by considering the interval $x \in \left(3 , \infty\right)$ that corresponds to $t \in \left(0 , \frac{\pi}{2}\right)$.

Then:

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

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - 4 x + 3}} = \int \frac{\sec t \tan t \mathrm{dt}}{\sqrt{{\sec}^{2} t - 1}}$

Use now the trigonometric identity:

${\sec}^{2} t - 1 = {\tan}^{2} t$

and as for $t \in \left(0 , \frac{\pi}{2}\right)$ the tangent is positive:

$\sqrt{{\sec}^{2} - 1} = \tan t$

so:

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - 4 x + 3}} = \int \frac{\sec t \tan t \mathrm{dt}}{\tan} t$

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - 4 x + 3}} = \int \sec t \mathrm{dt}$

The integral of $\sec t$ should be known, but here is how is calculated:

$\int \sec t \mathrm{dt} = \int \sec t \left(\frac{\sec t + \tan t}{\sec t + \tan t}\right) \mathrm{dt}$

$\int \sec t \mathrm{dt} = \int \left(\frac{{\sec}^{2} t + \tan t \sec t}{\sec t + \tan t}\right) \mathrm{dt}$

$\int \sec t \mathrm{dt} = \int \frac{d \left(\tan t + \sec t\right)}{\sec t + \tan t} \mathrm{dt}$

$\int \sec t \mathrm{dt} = \ln \left\mid \sec t + \tan t \right\mid + C$

Then:

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - 4 x + 3}} = \ln \left\mid \sec t + \tan t \right\mid + C$

and to undo the substitution we note that:

$\sec t = \left(x - 2\right)$

$\tan t = \sqrt{{\sec}^{2} t - 1} = \sqrt{{\left(x - 2\right)}^{2} - 1} = \sqrt{{x}^{2} - 4 x + 3}$

so:

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - 4 x + 3}} = \ln \left\mid x - 2 + \sqrt{{x}^{2} - 4 x + 3} \right\mid + C$

and by direct verification we can check that the solution is valid also for $x \in \left(- \infty , 1\right)$.