Int (1 Viewer)

[azureus88]

[maths]\int_{e}^{e^4}\frac{dx}{x\ln x}[/maths]

for the above question you could easily use the substitution x=lnx but if we were to use parts, we get:

[maths]2\int_{e}^{e^4}\frac{dx}{x\ln x}=[\frac{\ln x}{\ln x}]^{e^4}_{e}[/maths]

by letting [maths]u=\frac{1}{\ln x}[/maths] and [maths]v=\ln xdx[/maths]

Whats with that? How can you have 1 in the square brackets?

 

[Iruka]

You have lost a minus sign when you differentiated u.

All you have proven is that 0=0.

 

[azureus88]

yeh i did, thanks for pointing it out.

 

[Drongoski]

yeh i did, thanks for pointing it out.

The result is [ln(ln x))] (with limits e & e^4)

= (ln(ln e^4)) - (ln (ln e))

= ln 4 - ln 1

= ln 4

Is that right ????

 

[Timothy.Siu]

yeah thats right but he didn't want the answer lol

coincidentally i did exactly this question today

 

[tommykins]

Nvm :d

 

[Drongoski]

I don't think your solution is correct. If I'm right, the result of the integration is: ln(ln x)) + C

 

[tommykins]

oh wait i see what i did wrong