# Thread: Intergration Help

1. ## Intergration Help

how do i intergrate:

int(e^x(sqrt(1+e^x))) u=1+e^x

and

int((x^2)(e^x))

2. Originally Posted by Stylis10
how do i intergrate:

int(e^x(sqrt(1+e^x))) u=1+e^x

Mr F says: u = 1 + e^x => du/dx = e^x => dx = du/e^x. So the integral becomes int e^x sqrt(u) du/e^x = int sqrt(u) du.

and

int((x^2)(e^x))

Mr F says: Use integration by parts TWICE. First time will leave you with an integral of the form int x e^x dx. Second time leaves you with an integral of the form int e^x dx.
..

3. ok i understand the second 1 but am still puzzled by the first 1

4. Originally Posted by Stylis10
ok i understand the second 1 but am still puzzled by the first 1
What part puzzles you?

5. the whole thing kind of

when i work it all out they following the steps you've shown i get the final answer as (x+e^x)^-1, but that isnt right, is it?

6. Originally Posted by Stylis10
the whole thing kind of

when i work it all out they following the steps you've shown i get the final answer as (x=e^x)^-1, but that isnt right, is it?
$\int \sqrt{u}\, du = \int u^{1/2} \, du = \frac{2}{3} u^{3/2} + C = \frac{2}{3} u\, u^{1/2} + C = \frac{2}{3} u \, \sqrt{u} + C$.

Now substitute back u = 1 + e^x.

I have no idea what you've done. The working you've done leading to your answer would no doubt reveal your error(s).

7. ahhh thank you, i see where i have made my mistake, i substituted u = 1+e^x too soon. thank you for the help

8. Originally Posted by Stylis10

how do i intergrate:

int(e^x(sqrt(1+e^x))) u=1+e^x
A more direct way is to set $u^2=1+e^x\implies2u\,du=e^x\,dx,$

$\int {e^x \sqrt {1 + e^x } \,dx} = 2\int {u^2 \,du} = \frac{2}
{3}u^3 + k.$

The rest follows.