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

2 Answers
Mar 26, 2017

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

Explanation:

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.

Mar 27, 2017

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

Explanation:

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 lnabs(x-2) - 1/4 ln abs(x+2) +C

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