# Evaluate the integral from 2 to 3: 6/(x^2-1) dxI've looked at similar questions and if it was general then I think it would be 18*ln ((x-1)/(x+1)) + C, but since it's on a definite integral, all...

Evaluate the integral from 2 to 3: 6/(x^2-1) dx

I've looked at similar questions and if it was general then I think it would be 18*ln ((x-1)/(x+1)) + C, but since it's on a definite integral, all the options I'm trying are not giving me the right answer, and I don't know what that answer is. I've tried over 20 different things and my brain is about to explode, help!

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You know that 6/(x^2-1)= 6/[(x-1)(x+1)]

Then if you write 6/[(x-1)(x+1)]=A/(x-1)+B/(x+1);

By solving the above you will get A=3 and B=-3

Therefore 2∫3 ( 6/(x^2-1)dx = 2∫3 [3/(x-1)-3/(x+1)]dx

= 3 {2∫3 [1/(x-1)-1/(x+1)]dx}

= 3{ 2∫3 1/(x-1)dx-2∫3 1/(x+1)dx}

= 3 {ln(x-1)-ln(x+1)} from 2 to 3

= 3 ln[(x-1)/(x+1)] from 2 to 3

= 3 {ln[(3-1)/(3+1)]-ln[(2-1)/(2+1)]}

= 3 [ln(2/4)-ln(1/3)]

= 3 [ln(0.5)-ln(1/3)]

= 3 ln (0.5/(1/3))

= 3 ln (0.5*3)

= 3 ln 1.5

= 3*0.4055

= **1.216**

You dont need to have constants like C in definite integral because once we put the limits it will cancel out.

Evaluate `int_2^3 6/(x^2-1)dx` :

Let `u=x,du=dx` . Temporarily ignoring the limits of integration we get:

`int 6/(x^2-1)=-6int(du)/(1-u^2)` (Factor out the constant after multiplying by -1 to get the denominator in the form `a^2-u^2`

Using the inverse hyperbolic tangent `tanh^(-1)` we have:

`int (du)/(a^2-u^2)=1/(2a)ln|(a+u)/(a-u)|` Plugging in for a and u we get:

`-6int (du)/(1-u^2)=-6[1/2 ln|(1+u)/(1-u)|]=-3(ln|1+u|-ln|1-u|)`

Replacing u with x and bringing back the limits of integration we get:

`-3(ln|1+x|-ln|1-x|)|_2^3=-3(ln4-ln2)-(-3)(ln3-ln1)`

`=-3ln2+3ln3=3ln(3/2)~~1.216`

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`int_2^3 6/(x^2-1) dx=ln(27/8)~~1.216`

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