My Math Forum Stuck on an improper integral problem

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 February 28th, 2013, 10:36 PM #1 Newbie   Joined: Feb 2013 Posts: 4 Thanks: 0 Stuck on an improper integral problem My friend and I are completely stuck on this problem. When I get to the point to take the limit as t->infinity it would come out to ln|infinity/infinity| and that doesn't "work" right? We are just really confused. If someone could help explain that would be great. My Work
 March 1st, 2013, 12:07 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 461 Math Focus: Calculus/ODEs Re: Stuck on an improper integral problem You have correctly derived: $\int_1\,^{\infty}\frac{1}{2x^2+x}\,dx=\lim_{t\to\i nfty}$$\ln\(\frac{t}{2t+1}$$-\ln$$\frac{1}{3}$$\)$ As $t\to\infty$ we observe that $\frac{t}{2t+1}\to\frac{1}{2}$ hence: $\int_1\,^{\infty}\frac{1}{2x^2+x}\,dx=\ln$$\frac{1 }{2}$$-\ln$$\frac{1}{3}$$=\ln$$\frac{3}{2}$$$
 March 1st, 2013, 08:27 AM #3 Newbie   Joined: Feb 2013 Posts: 4 Thanks: 0 Re: Stuck on an improper integral problem Thanks! Now that I see what you did it's like "duh!" I think the part that was confusing us was taking the limit as t->infinity of ln|t/(2t+1)|. Just to clarify, you used l'Hôpital's rule to get 1/2, correct?
 March 1st, 2013, 09:56 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 461 Math Focus: Calculus/ODEs Re: Stuck on an improper integral problem While you can use L'Hôpital's rule, you could also rewrite the expression by dividing each term in the numerator and denominator by t: $\lim_{t\to\infty}\frac{1}{2+\frac{1}{t}}=\frac{1}{ 2+0}=\frac{1}{2}$

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