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

Question

Emily Ammerman · Accepted Answer

#int dx/sqrt(x^2-4x+3) = ln abs(x-2+sqrt(x^2-4x+3)) +C#

Complete the square at the denominator:

#x^2-4x+3 = (x-2)^2 -1#

and subtitute:

#x-2 = sect#

#dx = tant sect dt#

Note also that:

#x^2-4x+3 = (x-3)(x-1)#

and the integrand function is defined for

#x^2-4x+3 >0#

that is for #x in (-oo,1) uu (3,+oo)#. We will start by considering the interval #x in (3,oo)# that corresponds to #t in (0,pi/2)#.

Then:

#int dx/sqrt(x^2-4x+3) = int dx/sqrt((x-2)^2 -1)#

#int dx/sqrt(x^2-4x+3) = int (sect tant dt)/sqrt(sec^2t -1)#

Use now the trigonometric identity:

#sec^2t -1 = tan^2t#

and as for #t in (0,pi/2)# the tangent is positive:

#sqrt(sec^2-1) = tant#

so:

#int dx/sqrt(x^2-4x+3) = int (sect tant dt)/tant#

#int dx/sqrt(x^2-4x+3) = int sectdt#

The integral of #sect# should be known, but here is how is calculated:

#int sectdt = int sect ((sect+tant)/(sect+tant))dt#

#int sectdt = int ((sec^2t+tantsect)/(sect+tant))dt#

#int sectdt = int (d(tant+sect))/(sect+tant)dt#

#int sectdt = ln abs(sect+tant) +C#

Then:

#int dx/sqrt(x^2-4x+3) = ln abs(sect+tant) +C#

and to undo the substitution we note that:

#sect = (x-2)#

#tant = sqrt(sec^2t-1) = sqrt((x-2)^2-1) = sqrt(x^2-4x+3)#

so:

#int dx/sqrt(x^2-4x+3) = ln abs(x-2+sqrt(x^2-4x+3)) +C#

and by direct verification we can check that the solution is valid also for #x in (-oo,1)#.

Scarlett Allison · Answer

undefined To integrate $ \frac{1}{\sqrt{x^2 - 4x + 3}} $ with respect to $ x $, you can use a trigonometric substitution method. Let $ x - 2 = \sqrt{3} \sec(\theta) $. Then solve for $ x $ in terms of $ \theta $. After substitution, simplify the integrand, perform the integration with respect to $ \theta $, and then resubstitute back to $ x $ after integration.