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# (lnx)/x watch

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1. How would you integrate (lnx)/x ?
2. (Original post by Aired)
How would you integrate (lnx)/x ?
By parts:

Let u = lnx
.....dv/dx = 1/x

and proceed as normal.
3. (Original post by V.P. Keys)
By parts:

Let u = lnx
.....dv/dx = 1/x

and proceed as normal.
So... du/dx = 1/x and v = lnx.

= (lnx)/x - ∫(lnx)/x

4. (Original post by Aired)
So... du/dx = 1/x and v = lnx.

= (lnx)/x - ∫(lnx)/x

That's one way. Now rearrange the equation to make the integral you want the subject of it.
5. (Original post by Neapolitan)
That's one way. Now rearrange the equation to make the integral you want the subject of it.
Sorry I don't understand what you meant...
6. Take the integral on the RHS over to the LHS, so you get two lots of it.

Incidentally, the first term on the RHS should be (lnx)2...
7. (Original post by Aired)
Sorry I don't understand what you meant...
∫lnx/x dx = (lnx)² - ∫lnx/x dx

So

2 ∫lnx/x dx = (lnx)²

∫lnx/x dx = (lnx)²/2

Plus a constant!
8. (Original post by Neapolitan)
∫lnx/x dx = (lnx)² - ∫lnx/x dx

So

2 ∫lnx/x dx = (lnx)²

∫lnx/x dx = (lnx)²/2

Plus a constant!
Oh! Thanks, I'm so stupid sometimes
9. (Original post by Aired)
How would you integrate (lnx)/x ?
Even easier.

Let u = lnx
=> du/dx = 1/x
=> du = dx/x

∫(lnx)/x dx = ∫u du = ½u² = ½(lnx)² + c

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