How do you evaluate the integral #int 1/(x^2-4)#?

Question

Emily Ammerman · Accepted Answer

#int dx/(x^2-4) = 1/4 ln abs((x-2)/(x+2)) +C#

We can also show how to solve with partial fractions:

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

#1/((x-2)(x+2)) = A/(x-2)+B/(x+2)#

#1/((x-2)(x+2)) = (A(x+2)+B(x-2))/((x-2)(x+2))#

#1= Ax+2A+Bx-2B#

#1= (A+B)x +2(A-B)#

#{(A+B =0),(2A-2B =1):}#

#{(A = -B ),(4A =1):}#

#{(A=1/4),(B=-1/4):}#

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

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

#int dx/(x^2-4) = 1/4 lnabs(x-2) - 1/4 ln abs(x+2) +C#

#int dx/(x^2-4) = 1/4 ln abs((x-2)/(x+2)) +C#

Charles Anctil · Answer

#1/4lnabs((x-2)/(x+2))+C#

This can be done with partial fractions, but this will be done with a trigonometric substitution. First, we need to add #dx# to the integral:

#I=int1/(x^2-4)dx#

Recall that #sec^2theta-1=tan^2theta#.

Because of that identity, let #x=2sectheta#.

Importantly, this implies that #x^2-4=4sec^2theta-4=4(sec^2theta-1)=4tan^2theta#.

Furthermore, #dx=2secthetatanthetad theta#.

Then:

#I=int(2secthetatantheta)/(4tan^2theta)d theta=1/2intsectheta/tanthetad theta=1/2intcscthetad theta#

Which is a commonly known integral:

#I=-1/2lnabs(csctheta+cottheta)#

From our original substitution #x=2sectheta# we see that #sectheta=x/2#. This is a right triangle where #2# is the side adjacent to #theta#, #x# is the hypotenuse and then the side opposite theta is #sqrt(4-x^2)#.

Then, #csctheta=x/sqrt(4-x^2)# and #cottheta=2/sqrt(4-x^2)#. So:

#I=-1/2lnabs(x/sqrt(4-x^2)+2/sqrt(4-x^2))=1/2lnabs(sqrt(4-x^2)/(x+2))#

Expanding further:

#I=1/4lnabs(4-x^2)-1/2lnabs(x+2)=1/4lnabs(2+x)+1/4lnabs(2-x)-1/2lnabs(x+2)#

#I=1/4lnabs(2-x)-1/4lnabs(x+2)#

We can combine these and switch the order of #2-x# to #x-2# since we're working with absolute values:

#I=1/4lnabs((x-2)/(x+2))+C#

This would be much quicker if done with partial fractions.

Emilia Aguilar · Answer

undefined The integral of 1/(x^2 - 4) can be evaluated using partial fraction decomposition. After decomposition, it becomes A/(x - 2) + B/(x + 2), where A and B are constants to be determined. Then, integrate each term separately to obtain the final result.