Question

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

Answer 1

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

Explanation:

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`

Answer 2

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

Explanation:

`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`