1. ## stuck on integral

my teacher has given us this integration

$\displaystyle \int_0^x \frac{2. \lambda exp(- \lambda.x}{(1+exp(- \lambda.x)^2} dx$

I dont get how this equals

$\displaystyle \int_\frac{1}{2}^{\frac{1}{(1+exp(- \lambda.x)}} 2 du$

the domain of integration can change? can anyone explain this me?

2. Your equations seem to have a few errors. I think what you meant is this:

$\displaystyle \int _ 0 ^ X \frac {2 \lambda e ^{- \lambda x}} {(1 + e ^{- \lambda x} ) ^2 } dx$

Is that right? You are changing the variable from x to u using:

$\displaystyle u = \frac 1 {1 + e^ {-\lambda x}}$

From this you get:
$\displaystyle x = \frac {-1} {\lambda} ln( \frac {1-u} u)$, and $\displaystyle dx = \frac {1} {\lambda u (1-u)} du$

When you do this you have to change the limits of integration to conform to this new variable. So where initially you had a lower limit of x = 0, you now have $\displaystyle u = \frac 1 {1 + e^0} = \frac 1 2$. Likewise, the upper end becomes $\displaystyle u = \frac 1 {1 + e ^ {- \lambda X}}$. And the new integral is:

$\displaystyle 2\int _ {\frac 1 2 } ^ \frac 1 {1 + e ^ {- \lambda X}} du$

3. Originally Posted by notgoodatmath
my teacher has given us this integration

$\displaystyle \int_0^x \frac{2. \lambda exp(- \lambda.x}{(1+exp(- \lambda.x)^2} dx$

I dont get how this equals

$\displaystyle \int_\frac{1}{2}^{\frac{1}{(1+exp(- \lambda.x)}} 2 du$

the domain of integration can change? can anyone explain this me?

How the integration variable appears as upper limit in your integral??

Tonio

4. I think it's a typo. The OP used a lower case x both as his variable and to define the upper limit. In my response I used upper case X (not lower case x) in the upper llimit, so as to not confuse the two.