Help with integrating factor? I am currently solving a differential equation but I am having a little trouble figuring out my integrative factor. 
I have 
$$\exp\bigg({∫\frac{3}{t}dt}\bigg)$$
so to integrate I made it $\exp(\ln(t^3))$ what is the final solution? $t^3$?
 A: Here are the steps 
$$ \exp\left(\int\frac{3}{t}dt\right) = \exp\left(3\int\frac{1}{t}dt\right)$$
$$ = \exp\left(3\ln t\right) = \exp\left(\ln t^3\right) = t^3 $$
Since this is just an integrating factor, we do not need the constant. 
A: The answer to your confusion is that $\exp(x)=e^x$ and $\ln(x)=\log_e(x)$ are inverse functions.
This means that if you do one, and then the other, the action of the first is undone by the second. For example, $\ln e^2=2$ and $\exp(\ln(e^2))=e^2$.
You have to be careful though... you could put a negative number into $\exp(x)$ but not into $\ln x$...
Other examples of inverse functions include


*

*$f(x)=2x$ and $f^{-1}(x)=\frac{x}{2}$

*$g(x)=x^2$ and $g^{-1}(x)=\sqrt{x}$ (careful!)

*$h(x)=x+42$ and $h^{-1}(x)=x-42$.
In your case there is an assumption that $t>0$ and when you take 
$$t^3\longrightarrow \ln(t^3),$$
the action of $\exp(x)$ is to undo this and bring you back to $t^3$:
$$t^3\overset{\ln x}{\longrightarrow} \ln (t^3)\overset{\exp(x)}{\longrightarrow} t^3.$$
We have 
$$\exp(\ln x)=x\,,\qquad\text{for }x>0$$
and
$$\ln(\exp x)=x,$$
for any real $x$.
