# stuck on integral

#### 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?

#### ebaines

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:

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

#### tonio

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

#### ebaines

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.