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# ln 2x? watch

1. Can someone please explain to me how you integrate 2/2x? Because from what I can see you can put it as ln x or as ln 2x, but if you put limits in for this, you'll get completely different answers!
2. You simplify it to 1/x. I can't remember the reasoning behind why they produce different answers.
3. firstly 2/2x = 1/x.

lnx differentiates to 1/x
ln2x differetiates to 2/2x = 1/x

1/x can be intergrated into any lnax function where a is a positive constant
4. But if you put it as ln a x, isn't your integral value when you put limits in multiplied by a?
5. Well, ln(2x) = ln(2) + ln(x)

And if you put in limits for that, then the constant, ln(2), will just disappear.

You shouldn't have gotten different answers.
6. You still get an extra ln2 though!
7. The integral of 2/2x is lnx.

2/2x is the same as 1/x. The integral of 1/x is also lnx.
8. (Original post by monkee)
But if you put it as ln a x, isn't your integral value when you put limits in multiplied by a?
Well, yes, but BOTH limits are multiplied by a. And with logs, remember that ln(ax)= lna+lnx. So let's so your limits are u, v. ln(au)-ln(av)=
ln(u)+ln(a)-ln(v)-ln(a)=ln(u)-ln(v). So a doesn't matter.
9. (Original post by monkee)
But if you put it as ln a x, isn't your integral value when you put limits in multiplied by a?
try this,

ln(2x) [a .. b] = ln(2b) - ln(2a) = ln(2b/2a) = ln(b/a)
ln(x) [a .. b] = ln(b) - ln(a) = ln(b/a)
10. Oh, of course! Silly me! Thanks! Have my Core 3 exam tomorrow and a few last minute things popped into my head! I didn't think it made sense that they were equal, yet didn't produce equal integral answers...
11. The top is the differential of the bottom.
So the integral is ln2x.
12. (Original post by HitTheSwitch)
The top is the Differential of the bottom.
So the Integral is Ln2x.
The integral of 2/2x is not ln2x.

Or are you saying something else?
13. (Original post by Copacetic)
The integral of 2/2x is not ln2x.

Or are you saying something else?
The integral of 2/2x is ln2x + C; it's also lnx + D, where D = ln2 + C. The constants of integration are different, but that's the only difference. The OP's reasoning is sound, though a little roundabout.
14. (Original post by monkee)
You still get an extra ln2 though!
When you put limits, the ln2's will cancel each other out.
15. (Original post by Copacetic)
The integral of 2/2x is not ln2x.

Or are you saying something else?
The integral of 2/2x is ln|2x| + c.

But because earlier in the thread they were discussing limits, I assumed the + c was not needed.
16. (Original post by monkee)
Can someone please explain to me how you integrate 2/2x? Because from what I can see you can put it as ln x or as ln 2x, but if you put limits in for this, you'll get completely different answers!
2/2x = 1/x

integrate 1/x = ln(x) + c [just like any integration]

c = ln(a) [this is just a constant from some other function]

therefore integral = ln(x) + ln(a) = ln(ax), and a in this case is just 2.
17. .
or
are arbitrary constants. Hope this helps.

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Updated: January 19, 2010
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