Can you factor out a derivative inside an integral? I want to know if the following operation is legit and if no, why not? 
$$\text Let\ u= f(x) \ and\ v=g(x) $$ 
$$
\int uv' \ =?  \ \  v\int u
$$
Basically, can I take the derivative out? (Since integration is opposite to defferentiation)
Or perhaps that is not legit due to f(x) and g(x) being to different funcitons, can I do it if is the same function? 
$$
\int uu' \ =?  \ \  u\int u
$$
 A: Well, you can always just try some functions to see if it works. What happens if you let $f(x) = x$ and $g(x) = x$?
In general, it is true that
\begin{equation*}
 \int uv' \,\mathrm dx = uv - \int u'v\,\mathrm dx
\end{equation*}
(this is called integration by parts, and it is basically just a consequence of the product rule for differentiation but going the other way) and also that
\begin{equation*}
 \int uu'\,\mathrm dx = \tfrac 12 u^2 + C
\end{equation*}
which you can verify by differentiation.
A: No. You could do that for $\int_a^b v'(x) dx\ = v(x)|_{a}^{b}$
But not if there is another value in the integral. 
Counter example : $\int x \cdot ln'(x) dx = \int x \cdot \frac{1}{x} dx =\int 1 dx  \neq ln(x)\cdot\int x dx $
You might want to take a look at integration by parts
A: I presume the first isn't the case, since if we had an integral like $$ \int xe^x$$ 
we would need to use the chain rule and we can't simply take out $$ e^x $$
in front of the integral. But is there any operation we can do to factor out a derivative? 
A: No.  What you do have, with integration by parts, is: $\int uv'=uv-\int u'v$.
So for a counterexample, we just need $v\int u\ne uv-\int u'v$.  For instance, take $u=1,v=x$.
