Definite integral integration by parts Can we write the integration by parts for definite integral the following way:
$$\int^a_b f(x)g(x)dx=f(x)\int^a_b g(x)dx-\int^a_b  \left[  \dfrac{df(x)}{dx}\int^a_b g(x)dx   \right]dx   $$
My book gives the following formula for definite integral integration by parts:
$$\int^a_b f(x)g(x)dx=\left[f(x)\int g(x)dx\right]^a_b                  
-\int^a_b \left[  \dfrac{df(x)}{dx}\int g(x)dx  \right]dx  $$
Are the two formulas equivalent or not? Why/Why not?

Integration by parts is defined by
$$\int f(x) \, g(x) \, dx = f(x) \int g(u) \, du - \int f'(t) \left(\int^{t} g(u) \, du \right) \, dt.$$
When applying limits on the integrals they follow the form
$$\int_{a}^{b} f(x) \, g(x) \, dx = \left[f(x) \int g(u) \, du\right]_{a}^{b} - \int_{a}^{b} f'(t) \left(\int^{t} g(u) \, du \right) \, dt.$$
Now, if 
$$\int_{a}^{b} f(x) \, g(x) \, dx = f(x) \int_{a}^{b} g(u) \, du - \int_{a}^{b} f'(t) \left(\int_{a}^{b} g(u) \, du \right) \, dt$$
then what has been descried is
$$\int_{a}^{b} g(u) \, du$$
is a constant, say $c_{1}$ for which the remaining integration becomes
$$\int_{a}^{b} f(x) \, g(x) \, dx = c_{1} \, f(x) - c_{1} \, \int_{a}^{b} f'(t) \, dt.$$
These resulting integrals are not the same, in any sense, unless $g(x)$ is a constant to begin with.
As a demonstration consider $f(x) = x, g(x) =1$ for which the proper way yields
\begin{align}
\int_{1}^{2} f(x) \, g(x) \, dx &= \left[ x \cdot x \right]_{1}^{2} - \int_{1}^{2} 1 \cdot x \, dx \\
&= (4 - 1) - \left[ \frac{x^2}{2}\right]_{1}^{2} = \frac{3}{2}. 
\end{align}
The questioned method leads to
\begin{align}
\int_{1}^{2} f(x) \, g(x) \, dx &= x \int_{1}^{2} du - \int_{1}^{2} 1 \cdot \left(\int_{1}^{2} du \right) \, dt \\
&= x - \int_{1}^{2} dt \\
&= x - 1.
\end{align}
From this example even if it had been asked for
$$\int_{a}^{b} f(x) \, g(x) \, dx = \left[f(x) \int_{a}^{b} g(u) \, du\right]_{a}^{b} - \int_{a}^{b} f'(t) \left(\int_{a}^{b} g(u) \, du \right) \, dt$$
then the exampled result would be
\begin{align}
\int_{1}^{2} f(x) \, g(x) \, dx &= \left[x \int_{1}^{2} du\right]_{1}^{2} - \int_{1}^{2} 1 \cdot \left(\int_{1}^{2} du \right) \, dt \\
&= [x]_{1}^{2} - \int_{1}^{2} dt \\
&= 1 - 1 = 0.
\end{align}
which still leads to an incorrect result.
