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# Proof for 0=1 watch

1. Consider the integral of (1/xlnx).

I know that it is solved by using substitution u=lnx but consider solving via intergration by parts.

Let u=1/ln x and v'=1/x hence u'=-1/(x(lnx)^2) and v=ln x
So integral = uv- integral u'v which is:

lnx/lnx - (integral (-lnx/(x(lnx)^2)))

Which is:

1+ (integral (1/(xlnx)) which is the integral of (1/(xlnx)).

So if we let I= the original integral we get

I= 1+I or 0=1.

What have I done wrong?
2. (Original post by Econowizard)
Consider the integral of (1/xlnx).

I know that it is solved by using substitution u=lnx but consider solving via intergration by parts.

Let u=1/ln x and v'=1/x hence u'=-1/(x(lnx)^2) and v=ln x
So integral = uv- integral u'v which is:

lnx/lnx - (integral (-lnx/(x(lnx)^2)))

Which is:

1+ (integral (1/(xlnx)) which is the integral of (1/(xlnx)).

So if we let I= the original integral we get

I= 1+I or 0=1.

What have I done wrong?
it is probably due to not using the modulus when you integrated 1/x
3. (Original post by the bear)
it is probably due to not using the modulus when you integrated 1/x
How does that affect anything?
4. You forgot the constant I guess?
5. (Original post by kishen111)
You forgot the constant I guess?
Thats what it probably is but consider if you set limits a,b. Then no constant would be produced and they would cancel out perfectly?
6. (Original post by Econowizard)
Thats what it probably is but consider if you set limits a,b. Then no constant would be produced and they would cancel out perfectly?
With limits you get

7. (Original post by Econowizard)
Thats what it probably is but consider if you set limits a,b. Then no constant would be produced and they would cancel out perfectly?
Check this out for the indefinite integral.

https://math.stackexchange.com/quest...371652#1371652
8. (Original post by RDKGames)
With limits you get

Awesome thank you!

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