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?

2 Answers
May 16, 2017

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.

I use to solve this way:
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