# Math Help - limit of a proper integral

1. ## limit of a proper integral

How would you write:
intergral from 2 to infinity of (x+3)/[(x^2-1)(x-1)]

as the limit of a proper intergral, then evaluate the intergral.
Then take the limit to find the improper definite intergral?

The partial fraction is (1/2)ln(x+1) - (1/2)ln(x-1) - 2/(x-1)
thanks

2. Well you can set

$\int_2^\infty\frac{x+3}{(x^2-1)(x-1)}\,dx=\lim_{a\to\infty}\int_2^a\frac{x+3}{(x^2-1)(x-1)}\,dx$

Then solve.

3. Yea i get that however i am unsure what to do when it comes out as
[1/2ln(a+1)-2/(a-1)-1/2ln(a-1)]-[1/ln(3)-2]
what do you do with the ln terms do you write ln(a+1/a-1)
Then what do you do?
Thanks

4. $\mathop {\lim }\limits_{a \to \infty } \int_2^a {\frac{{x + 3}}
{{\left( {x^2 - 1} \right)\left( {x - 1} \right)}}\,dx} = \mathop {\lim }\limits_{a \to \infty } \left( {\left. { - \frac{1}
{2}\ln \left| {x - 1} \right| + \frac{1}
{2}\ln \left| {x + 1} \right| - \frac{2}
{{x - 1}}} \right|_2^a } \right)$

So, when you're evaluating for $a$, turn back into $\infty$

5. But then we get ln(infinity) which is infinty??

6. Hello, Ben3535!

It seems that you found the correct partial fractions and you integrated correctly.

So what left?

How would you write: intergral from 2 to infinity of (x+3)/[(x^2-1)(x-1)]

as the limit of a proper intergral, then evaluate the intergral.
Then take the limit to find the improper definite intergral?

The partial fraction is (1/2)ln(x+1) - (1/2)ln(x-1) - 2/(x-1)

Partial fractions: . $\frac{x+3}{(x+1)(x-1)^2} \;=\;\frac{\frac{1}{2}}{x+1} + \frac{\text{-}\frac{1}{2}}{x-1} + \frac{2}{(x-1)^2}$

Integrate: . $\frac{1}{2}\int\frac{dx}{x+1} - \frac{1}{2}\int\frac{dx}{x-1} + 2\int(x-1)^{-2}dx$

. . $=\;\frac{1}{2}\ln(x+1) - \frac{1}{2}\ln(x-1) - \frac{2}{x-1} \;=\;\frac{1}{2}\ln\left(\frac{x+1}{x-1}\right) - \frac{2}{x=1}$

We will evaluate from 2 to $\infty$

. . $\lim_{b\to\infty}\bigg[\frac{1}{2}\ln\left(\frac{x+1}{x-1}\right) - \frac{2}{x-1}\bigg]^b_2 \;= \;\lim_{b\to\infty}\bigg[\left(\frac{1}{2}\ln\frac{b+1}{b-1} - \frac{2}{b-1}\right) - \left(\frac{1}{2}\ln\left(\frac{3}{1}\right) - \frac{2}{1}\right)\bigg]$

In the first fraction, divide top and bottom by $b$.

. . $\lim_{b\to\infty}\bigg[\frac{1}{2}\ln\left(\frac{1 + \frac{1}{b}}{1 - \frac{1}{b}}\right) - \frac{2}{b-1} - \frac{1}{2}\ln(3) + 2\bigg]$

. . $=\;\frac{1}{2}\ln\left(\frac{1+0}{1-0}\right) - 0 - \frac{1}{2}\ln(3) + 2$

. . $= \;\frac{1}{2}\ln(1) - \frac{1}{2}\ln(3) + 2 \;=\;\boxed{2 - \frac{1}{2}\ln(3)}$