How Do You Solve the Integral of 2/(x^2-1)dx?

[masonm127]

Integral of 2/(x^2-1)dx HELP!

Homework Statement

Evaluate: ∫2/(x²-1)dx

Homework Equations

Inverse trig functions arent working.
∫1/x+1dx=ln(1+x)

The Attempt at a Solution

My tutor and i both attemped the solution many ways, and are still at a stand still.
According to my calculator and wolfram alpha, the answer is -ln(x+1/x-1)
The closest i have gotten is 2∫$1/((x+1)(x-1))$dx

 

[Dick]

Did you try using partial fractions?

 

[ArcanaNoir]

if you multiply by -1 I think you can use tanh^-1

 

[masonm127]

Thanks! partial fractions worked! didn't even think about that. the solution came out to be -ln(x+1)+ln(x-1), which is apparently equal to -ln((x+1)/(x-1))

 

[ArcanaNoir]

and also apparently equal to tanh^-1

 

[masonm127]

The full question is to find the integral over infinity and 2, so i have to use limits. I am getting an undefined answer since ln of infinity minus ln infinity is undefined. does that sound right?

 

[Dick]

The full question is to find the integral over infinity and 2, so i have to use limits. I am getting an undefined answer since ln of infinity minus ln infinity is undefined. does that sound right?

Not right. You should really be writing things like the integral of 1/(1+x) as log(|1+x|). Since it's an improper integral, to work out the infinity part of your limit you need to think about the limit as x->infinity of log(|1+x|/|1-x|). What's that?

 

[masonm127]

should be zero right? Not used to using log to write things, its just something our teacher never really told us to do. makes sense though

 

[Mark44]

Although my choice would be partial fractions decomposition, a trig substitution will also work here, with secθ = x, secθtanθdθ = dx. The resulting integral is
$$2\int csc\theta d\theta$$