I'm doing integrating by parts, and I came across a question that gave dv= e^{r/2}
Why is the anti-derivative 2e^{r/2 }?
I know my work is wrong for what I have, and I'm really not sure why it is.
You're welcome! ^^
Uhm, I don't know what you actually wanted to ask, but yes, there is a rule:
$\displaystyle \left (e^{f(x)} \right )'=f'(x) \cdot e^{f(x)}$
Also, if $\displaystyle f(x)=ax+b$, then:
$\displaystyle \left (e^{ax+b} \right )'=a \cdot e^{ax+b}$
So the antiderivate of $\displaystyle e^{ax+b}$ is $\displaystyle \frac{1}{a}e^{ax+b}$.
Be careful about using "x" there since it will typically be interpreted as a variable and you mean it as a constant here. It is true that the derivative of e^(ar), with respect to r, where a is a constant, is ae^(ar) so that the derivative of e^(ar)/a, with respect to r, is ae^(ar)/a= e^(ar) so that the antiderivative of e^(ar) (again with constant a) is e^(ar)/a.