Does $\int \frac{1}{u} \frac{du}{dt} dt$ violate integration by parts? Given that $u = u(t)$, does $\int \frac{1}{u} \frac{du}{dt} dt$ violate integration by parts, since the equation below is not possible?
$$ \int \frac{1}{u} \frac{du}{dt} dt = \frac{1}{u} u - \int u \left(-\frac{1}{u^2} \frac{du}{dt} \right)dt$$
$$ \therefore 0 = 1$$
Also, 
$$\int f(u) \frac{du}{dt} dt = \int \frac{1}{2}\frac{d}{dt}[f(u)]^2\ dt = \frac{1}{2} [f(u)]^2$$
but if $f(u)=1/u$, the equation above gives
$$\int f(u) \frac{du}{dt} dt = \frac{1}{2u^2}$$
which is not true.
What is happening with this function?
 A: First, $\frac{du}{dt}dt=du$.  So, simplifying what you're writing comes to:
$$
\int \frac{1}{u}du.
$$
Then, you do integration by parts with $f=\frac{1}{u}$ and $g'=du$.  Then, $f'=-\frac{1}{u^2}du$ and $g=u$.  In this case, the integration by parts comes to
$$
\int\frac{1}{u}du=1+\int\frac{1}{u}du.
$$
Now, this looks problematic, but it can be resolved by supposing that $F(u)$ is an antiderivative of $\frac{1}{u}$.  In this case, the expression simplifies to
$$
(F(u)+C_1)=1+(F(u)+C_2)
$$
where $C_1$ and $C_2$ are constants of integration.  Therefore, $1=C_1-C_2$, which is not a problem.  The issue is that $\int\frac{1}{u}du$ can represent two different functions in the same equation.
Note that if you instead used a definite integral, then you would have
$$
\int_a^b\frac{1}{u}du=\left.1\right|_a^b+\int_a^b\frac{1}{u}du.
$$
Moreover $\left.1\right|_a^b=1-1=0$, so this simplifies to
$$
\int_a^b\frac{1}{u}du=\int_a^b\frac{1}{u}du.
$$
Once the $1$ vanishes, this might make you feel a bit more comfortable.
In the second example, there is a missing chain rule.
$$
\frac{1}{2}\frac{d}{dt}[f(u)]^2\not=f(u)\frac{du}{dt},
$$
but
$$
\frac{1}{2}\frac{d}{dt}[f(u)]^2=f(u)f'(u)\frac{du}{dt}.
$$
You're missing a derivative of $f$.
