Question

What is `int1/(x^2-5)`?

When I use an online integral calculator, it's suggesting I use substitution and let `u=x/sqrt(5)`

I don't understand how to get the `sqrt(5)`. Is there an alternative way of solving this?
If not - what does the `sqrt(5)` mean and how do I work this out for similar questions?

Answer 1

`ln abs((x-sqrt5)/(x+sqrt5))/(2sqrt5) +C`

Explanation:

`int 1/(x^2-5)=int (1/5)/((1/5)(x^2-5))=1/5 int 1/(x^2/5 -1)=1/5 int 1/((x/sqrt(5))^2 -1)`

Now if we set `u=x/sqrt(5)` then the numerator must be `1/sqrt5` since `(du)/(dx)[u]=1/sqrt(5)` to apply the substitition rule.

`1/(5(1/sqrt5)) int ((1/sqrt5)1)/((x/sqrt(5))^2 -1) dx`

It is now all set for substitution:
`sqrt(5)/5 int 1/(u^2-1)du=(sqrt(5)/5)[1/2(int 1/(u-1)du-int 1/(u+1) du)]=(1/(2sqrt(5))[ln(u-1)-ln(u+1)])=(ln[(u-1)/(u+1)])/(2sqrt5)`

Undo substitution:
`(ln[(x/sqrt5-1)/(x/sqrt5+1)])/(2sqrt5)=ln abs((x-sqrt5)/(x+sqrt5))/(2sqrt5) +C`

Absolute value for the domain.

Answer 2

I use to solve this way: