How do you integrate #1/(x^2 - 3)#?

Answer 1

#=1/(2sqrt3)lnabs((x-sqrt3)/(x+sqrt3))+C#

Decompose #1/(x^2-3)# using partial fractions:
#1/(x^2-3)=1/((x+sqrt3)(x-sqrt3))=A/(x+sqrt3)+B/(x-sqrt3)#
#1/(x^2-3)=(A(x-sqrt3)+B(x+sqrt3))/(x^2-3)#
#(0x+1)/(x^2-3)=(x(A+B)+(-sqrt3A+sqrt3B))/(x^2-3)#

Thus:

#{(A+B=0),(-sqrt3A+sqrt3B=1):}#
Multiplying the first equation by #sqrt3#:
#{(sqrt3A+sqrt3B=0),(-sqrt3A+sqrt3B=1):}#
Adding them gives #2sqrt3B=1# so #B=1/(2sqrt3)#.
Since #A+B=0#, we see that #A=-1/(2sqrt3)#.

Thus:

#intdx/(x^2-3)=-1/(2sqrt3)intdx/(x+sqrt3)+1/(2sqrt3)intdx/(x-sqrt3)#
#=-1/(2sqrt3)lnabs(x+sqrt3)+1/(2sqrt3)lnabs(x-sqrt3)#
#=(lnabs(x-sqrt3)-lnabs(x+sqrt3))/(2sqrt3)#
#=1/(2sqrt3)lnabs((x-sqrt3)/(x+sqrt3))+C#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

#1/(2sqrt3)lnabs((x-sqrt3)/(x+sqrt3))+C#

#I=intdx/(x^2-3)#
Let #x=sqrt3sectheta#, implying that #dx=sqrt3secthetatanthetad theta#:
#I=int(sqrt3secthetatanthetad theta)/(3sec^2theta-3)=sqrt3/3int(secthetatanthetad theta)/(sec^2theta-1)#
Note that #sec^2theta-1=tan^2theta#:
#I=1/sqrt3int(secthetatanthetad theta)/tan^2theta=1/sqrt3int(secthetad theta)/tantheta=1/sqrt3int1/costheta(costheta/sintheta)d theta=1/sqrt3intcscthetad theta#

This is a common integral:

#I=-1/sqrt3lnabs(csctheta+cottheta)#
Note that #sectheta=x/sqrt3#, thus we have a right triangle where #x# is the hypotenuse, #sqrt3# is the side adjacent to #theta#, and the opposite side is #sqrt(x^2-3)#.
Thus:#" "csctheta=x/sqrt(x^2-3)" "#and#" "cottheta=sqrt3/sqrt(x^2-3)#
#I=-1/sqrt3lnabs(x/sqrt(x^2-3)+sqrt3/sqrt(x^2-3))#
#I=-1/sqrt3lnabs((x+sqrt3)/sqrt(x^2-3))#
(Technical note. This almost could be a final answer, but there is one problem, which is that #sqrt(x^2-3)# in the answer restricts the domain, in that we see that the domain excludes #-sqrt3 < x < sqrt3#. This is not the case in the original function, but the trigonometry introduced this issue. To remedy this, add absolute value bars.)
#I=-1/sqrt3lnabs((x+sqrt3)/sqrt(abs(x^2-3)))#

This is a proper final answer. However, we can simplify it rather sneakily:

#I=-1/sqrt3lnabs(sqrt((x+sqrt3)^2/(abs(x^2-3)))#
The square root is a #1/2# power, which can be brought from the logarithm rule: log(A^B)=Blog(A)#.
#I=-1/(2sqrt3)lnabs((x+sqrt3)^2/(x^2-3))#
#I=-1/(2sqrt3)lnabs((x+sqrt3)^2/((x+sqrt3)(x-sqrt3)))#
#I=-1/(2sqrt3)lnabs((x+sqrt3)/(x-sqrt3))#
We can also use the previous rule in reverse: #Blog(A)=log(A^B)#, to bring the #-1# power in and flip the fraction:
#I=1/(2sqrt3)lnabs((x-sqrt3)/(x+sqrt3))+C#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 3

To integrate ( \frac{1}{{x^2 - 3}} ), you can use a technique called partial fraction decomposition. First, factor the denominator:

[ x^2 - 3 = (x - \sqrt{3})(x + \sqrt{3}) ]

Then, express ( \frac{1}{{x^2 - 3}} ) as the sum of two fractions:

[ \frac{1}{{x^2 - 3}} = \frac{A}{{x - \sqrt{3}}} + \frac{B}{{x + \sqrt{3}}} ]

Find the values of ( A ) and ( B ) by equating numerators:

[ 1 = A(x + \sqrt{3}) + B(x - \sqrt{3}) ]

Solve for ( A ) and ( B ), which yields ( A = \frac{1}{2\sqrt{3}} ) and ( B = -\frac{1}{2\sqrt{3}} ).

Now, integrate each fraction separately:

[ \int \frac{1}{{x^2 - 3}} , dx = \frac{1}{2\sqrt{3}} \ln|x - \sqrt{3}| - \frac{1}{2\sqrt{3}} \ln|x + \sqrt{3}| + C ]

Where ( C ) is the constant of integration.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7