∫dx/x^2 - 4

Question

∫dx/x^2 - 4

loser66 · Answer

x^2 -4= (x+2)(x-2) then partial fraction to take integral

anonymous · Answer

Loser66....shud we then bring it to the numerator??

loser66 · Answer

sure

anonymous · Answer

but then wont it hav power -1....so we cant use the formula int x^n=x^n+1/n+1

loser66 · Answer

nope, you misunderstand my suggestion.
let integral symbol aside.
the integrand is $$\frac{1}{(x+2)(x-2)}= \frac{A}{x+2}+\frac{B}{x-2}
solve ~for~ A ~and~B, $$ 
then plug back to the integral to calculate
It's called integral by partial fraction method

loser66 · Answer

someone helps me explain him, please

loser66 · Answer

I don't know, I don't calculate it, just give you the method.

.sam. · Answer

The answer is not ln|x+2|+ln|x-2|+c, that is just an example by cambrige

loser66 · Answer

@cambrige  Sorry, I don't understand what you wrote "sum1 tell if I'm r8". English is not my  native language.

anonymous · Answer

*meant

loser66 · Answer

to me,  the answer is $$\frac{-1}{4}ln (x+2) +\frac{1}{4}ln(x-2)+ C$$

anonymous · Answer

The answer in my book is 1/4ln[x-2/x+2]+c

loser66 · Answer

the same, just another way to write ln

anonymous · Answer

Thanks a lot

anonymous · Answer

so I shud delete all the comments??

.sam. · Answer

@cambrige No you need not to