Improper integral and L'Hôpital's rule

[rayne1]

integral from 2 to infinity dx/(x^2+2x-3)

I got this as the result:
lim x to infinity (1/4)(ln|x-1|-ln|x+1|+ln|5|)

Then I got (1/4)(infinity - infinity + ln|5|) so do I need to use l'hopital's rule for ln|x-1|-ln|x+1| or would the final answer be ln|5|/4? If not, I am unsure of how to l'hopital's rule for this kind of question.

 

[MarkFL]

We are given to evaluate:

$$I=\int_2^{\infty}\frac{1}{x^2+2x-3}\,dx$$

As this is an improper integral, we should write:

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

Factoring the denominator of the integrand and applying partial fraction decomposition, we obtain:

$$I=\frac{1}{4}\lim_{t\to\infty}\left(\int_2^{t}\frac{1}{x-1}-\frac{1}{x+3}\,dx\right)$$

Applying the FTOC, we get:

$$I=\frac{1}{4}\lim_{t\to\infty}\left(\left[\ln\left|\frac{x-1}{x+3}\right|\right]_2^{t}\right)$$

$$I=\frac{1}{4}\lim_{t\to\infty}\left(\ln\left|\frac{t-1}{t+3}\right|+\ln(5)\right)=\frac{1}{4}\left(\ln\left(\lim_{t\to\infty}\left(\frac{t-1}{t+3}\right)\right)+\ln(5)\right)=\frac{\ln(5)}{4}$$

Your solution is correct, and there is no need to use L'Hôpital's rule here.