# Thread: just need explanation for integration prob.

1. ## just need explanation for integration prob.

Ok so this is the problem:

Integral from 1 to e of (ln(x))/x) dx (sorry don't know how to put the intergral symbol in)

I set u= lnx and du= 1/x and changed the limits to 0 to 1
When i solve though, I get 1, and my book says the answer is 1/2. I've tried this a bunch of times but can't get their answer. Any explanation?

2. Originally Posted by painterchica16
Ok so this is the problem:

Integral from 1 to e of (ln(x))/x) dx (sorry don't know how to put the intergral symbol in)

I set u= lnx and du= 1/x and changed the limits to 0 to 1
When i solve though, I get 1, and my book says the answer is 1/2. I've tried this a bunch of times but can't get their answer. Any explanation?
So after your sub you get

$\displaystyle \int_{0}^{1}u du =\frac{1}{2}u^2 \bigg|_{0}^{1}=\frac{1}{2}(1^2-0^2)=\frac{1}{2}$

3. $\displaystyle \int_{1}^{e} \frac{lnx}{x}$

$\displaystyle u = lnx$

$\displaystyle du = 1/x$

$\displaystyle \int udu$

Raise power and divide by power.

$\displaystyle \int udu = \frac{1}{2}u^2$

$\displaystyle \frac{(lnx)^2}{2}$

From 1 to e,

1/2ln(e)^2 - 1/2ln(1)^2

ln(1) = 0, ln(e) = 1

4. Thanks, I understand the setup now, but the only that I'm still confused about is why you plugged 0 and 1 into just u^2, and not lnx?

Because, ln(1)^2=0 and ln(0)^2 just doesnt exist. Why does it not work when you plug it in?

5. If u = lnx, than our intervals are no longer 1 and e but ln(1) and ln(e). Hence the substitution.

6. Oh, I get it, you plug it into u^2, not the other thing, because its all in terms of u. Thanks!