1. ## Integral problem

Hi guys i've got a problem with an integral question:

2. Hello,
Originally Posted by Redeemer_Pie
Hi guys i've got a problem with an integral question:

This can be rewritten :
$\int t \cdot 2^{-2t} ~dt=\int t \left[e^{\ln(2)}\right]^{-2t} ~dt=\int t e^{-2t \ln(2)} ~dt=\int t e^{at} ~dt$

where $a=-2\ln(2)$

and now, do an integration by parts

3. thanks alot!!! :3

4. I was wondering is the final answer:

t(2^(-2t))/(-2ln2t) - 2^(-2t)/(-2ln2t)^2

can someone confirm for me? It doesnt look quite right.

thanks guys

5. Originally Posted by Redeemer_Pie
I was wondering is the final answer:

t(2^(-2t))/(-2ln2t) - 2^(-2t)/(-2ln2t)^2

can someone confirm for me? It doesnt look quite right.

thanks guys
What you have typed is almost impossible to decipher, it should be:

$-{{\left(2\,\ln(2)\,t+1\right)\,2^{-2t}}\over{4\,(\ln(2))^2}}$

CB

6. my apologies is that the final answer?. how did you work that out?

t 'multiply' (e^(-2ln2t)/(-2ln2t)) - 'integral sign' (e^(-2ln2t)/(-2ln2t))

Thanks again

7. Originally Posted by Moo
Hello,

This can be rewritten :
$\int t \cdot 2^{-2t} ~dt=\int t \left[e^{\ln(2)}\right]^{-2t} ~dt=\int t e^{-2t \ln(2)} ~dt=\int t e^{at} ~dt$

where $a=-2\ln(2)$

and now, do an integration by parts
Working from here, we have $u=t$ and $dv=e^{at}\,dt$. Thus $du = dt$ and $v = \frac{1}{a}e^{at}$. Now we have:

$\frac{t}{a}e^{at}-\frac{1}{a}\int e^{at} = \frac{t}{a}e^{at}-\frac{1}{a^2}e^{at} = \frac{e^{at}}{a^2}(at-1)$

Since $a=-2\ln(2)$ and $e^{-2t\ln(2)} = 2^{-2t}$, we have:

$\frac{e^{at}}{a^2}(at-1) = \frac{2^{-2t}(-2t\ln(2)-1)}{4(\ln(2))^2} = \boxed{-\frac{2^{-2t}(2t\ln(2)+1)}{4(\ln(2))^2}}$

8. thanks a lot man.