# Thread: Can't work out why u=e^x means u^2 becomes e^x ?

1. ## Can't work out why u=e^x means u^2 becomes e^x ?

Question says Show that the substitution $\displaystyle u = e^x$ converts

$\displaystyle \int \frac{2 + ln u}{u^2} du$ into

$\displaystyle \int \frac{2 + x}{e^e} dx$

I understand how ln e^x = xlne = x .

But Why does u^2 become e^x ? Would it become e^x squared?

Angus

2. ## Re: Can't work out why u=e^x means u^2 becomes e^x ?

Hi angypangy!

Yes, $\displaystyle u^2$ becomes $\displaystyle (e^x)^2$.

But $\displaystyle du$ becomes $\displaystyle e^x dx$.

The $\displaystyle e^x$ cancels against the denominator.

3. ## Re: Can't work out why u=e^x means u^2 becomes e^x ?

Originally Posted by angypangy
Question says Show that the substitution $\displaystyle u = e^x$ converts

$\displaystyle \int \frac{2 + ln u}{u^2} du$ into

$\displaystyle \int \frac{2 + x}{e^e} dx$

I understand how ln e^x = xlne = x .

But Why does u^2 become e^x ? Would it become e^x squared?

Angus

From the laws of logarithms $\displaystyle \ln(e^x)=x \ln(e)$, and by definition $\displaystyle \ln(e)=1$

CB

4. ## Re: Can't work out why u=e^x means u^2 becomes e^x ?

Originally Posted by ILikeSerena
Hi angypangy!

Yes, $\displaystyle u^2$ becomes $\displaystyle (e^x)^2$.

But $\displaystyle du$ becomes $\displaystyle e^x dx$.

The $\displaystyle e^x$ cancels against the denominator.

I am not getting this. Could you please work through how to do it.

This is where I get to.

$\displaystyle \int \frac{2 + x}{e^e} dx$

Then let u = e^x

$\displaystyle \frac{du}{dx} (u = e^x) = e^x$

ie du = e^x dx

$\displaystyle \int \frac{(2 + x)(u)}{u} du$

and

Then what?

5. ## Re: Can't work out why u=e^x means u^2 becomes e^x ?

Originally Posted by angypangy
I am not getting this. Could you please work through how to do it.

This is where I get to.

$\displaystyle \int \frac{2 + x}{e^e} dx$

Then let u = e^x

$\displaystyle \frac{du}{dx} (u = e^x) = e^x$

ie du = e^x dx

$\displaystyle \int \frac{(2 + x)(u)}{u} du$

and

Then what?
Starting from: du = e^x dx

This means $\displaystyle dx = {du \over e^x} = {du \over u}$

So:

$\displaystyle \int \frac{(2 + x)}{u} {du \over u} = \int \frac{(2 + \ln u)}{u^2} du$