Question

Evaluating an integral by parts. Question about setting it up. Give me a second to put it below.

Answer 1

\[\lim_{b \rightarrow \infty}(-\frac{ x }{ x+3 } + \ln|x+3|)\]
This is how far I have got, the original problem is not there. Also I couldn't figure out the notation, but both of those parts are from 2 to b.

Answer 2

Normally to evaluate one statement I plug in the b to the function, then subtract the function with the 2 in it this time, but this one has 2 functions so I am confused on what to do.

Answer 3

Where's the original problem? :| can we see that..?

Answer 4

Yes, \[\int\limits_{2}^{\infty} \frac{ x }{ (x+3)^2 } dx\]

Answer 5

I did integration by parts and that led me to the evaluating part. Normally when I started doing these problems I would just have maybe the ln|x+3| by itself and plugging in was easy, now I don't get how to plug in the numbers.

Answer 6

Ok your setup looks pretty close.
So here is the upper boundary, as you described:
\[\large \lim_{b \rightarrow \infty}\left(-\frac{b}{b+3}+\ln|b+3|\right)\]And we want to subtract from that, the function evaluated at the lower limit,\[\large \large \lim_{b \rightarrow \infty}\left(-\frac{b}{b+3}+\ln|b+3|\right)-\left(\color{royalblue}{-\frac{2}{2+3}+\ln|2+3|}\right)\]

Answer 7

Yah, we can rewrite that first part as the sum of limits, allowing us to evaluate each limit separately. That might make it a little easier to understand.\[\large \large \lim_{b \rightarrow \infty}\frac{-b}{b+3}+\lim_{b \rightarrow \infty}\ln|b+3|-\left(\color{royalblue}{-\frac{2}{2+3}+\ln|2+3|}\right)\]

Answer 8

So basically we substitute the top term (b) in for the whole thing, and then the bottom term (2) for the whole thing again, and it gets subtracted?

Answer 9

Yes :) Try to remember back to the fundamental theorem:\[\large \int\limits_a^b f(x)dx \qquad=\qquad F(b)-F(a)\]

Big F represents that whoooole thing.
And we're plugging \(\large b\) in for every x in our Big F, then subtracting the whoooole thing again, with a's plugged in.

Answer 10

That helps so much! Thank you!