Question

So stuck. :(

∫(1/x^2+x)dx

I'll post an image of my work below.

Answer 1

I don't know how many times I've tried this now...

Answer 2

Oops, I should qualify, that's from 1 to infinity.

Answer 3

I'm pretty confident until I get to the point where it's equal to the limit as b approaches INF of -2 times the integral from 1 to b of 1/(1-s^2) ds.

Answer 4

1/x^2+x = \(\dfrac{1}{x^{2}} + x\)

If you mean: 1/(x^2 + x) = \(\dfrac{1}{x^{2}+x}\) The parentheses are NOT optional.

Answer 5

I meant ∫(1/(x^2+x)).

Answer 6

Sorry about that. The stress had my typing a little sloppily. \[\int\limits_{1}^{\infty}\frac{ 1 }{x^{2}+ x}\]

Answer 7

Are you sure it converges?

Answer 8

Well, I'm *reasonably* sure. The problem reads "Find the area of the region under the curve" and then the equation.

Answer 9

Have you considered Partial Fractions? \(\dfrac{1}{x^{2} + x} = \dfrac{1}{x} - \dfrac{1}{x+1}\)

This may or may not discourage you.

Answer 10

Actually, that may make me slap my head and say, "Aw DUHH".

Answer 11

Maybe, but neither of those converges by itself. The difference is smaller than either piece by itself and actually does converge.

Answer 12

Hmm... so I'd take the limit of each as b -> 0?

Answer 13

Add them first. Then worry about the limit. You're going to have to come to terms with neither piece converging but the difference behaving otherwise. Throw in a logarithm property to finish up.

Answer 14

Ah...that's sounding better. Ok, I think I can do that.

Answer 15

BTW Amazing Work!!! You really hung in there and it appears you tried just about everything! Seriously, you probably got more value out of this problem than anyone!

Answer 16

Thanks. I was seriously freaking out. I'm about to quit my job in a day or two so I can do this full time, and this one was really starting to shake my confidence. :)

Answer 17

Don't lose that confidence!!! You've clearly got the hard work going for you. Keep it up.

Answer 18

Thanks, seriously. I'm going to take a short break for dinner and be social with the wife and kids, but I really appreciate it. I'll check back in as soon as I work through it.

Answer 19

Ok, it *looks* like I'm on the right path, but on the bottom of the image, I have lim b->inf of ln |1-1|. That doesn't seem right.

Answer 20

1) Everything is positive, so forget about the absolute values.
2) You can't have subtraction inside the logarithm

\(\int\limits_{1}^{b}\dfrac{1}{x} - \dfrac{1}{x+1}\;dx = ln(2)-ln(b+1)+ln(b) = ln(2) + ln\left(\dfrac{b}{b+1}\right)\)

Careful! Now, work on the limit.

Answer 21

Ah, yes! It helps to write the down the factors right. Ok, one more try.

Answer 22

Holy crap, I finally got it! Thanks again for your help. If I could afford it, I'd probably build a statue of your avatar! :)