To simplify the expression

\[
\frac{2a}{(a+1)^2} + \frac{a}{a+1} + 1 - \frac{a}{a^2} + 2a + \frac{12a}{(a+1)^2} + \frac{a}{a+1} + 1 - \frac{a}{a^2} + 2a + 1,
\]

we can start by combining like terms. Let's rewrite the expression clearly, grouping similar terms together:

1. First, combine the fractions having \((a+1)^2\) in the denominator:

\[
\frac{2a}{(a+1)^2} + \frac{12a}{(a+1)^2} = \frac{(2a + 12a)}{(a+1)^2} = \frac{14a}{(a+1)^2}.
\]

2. Next, combine the terms with \(\frac{a}{a+1}\):

\[
\frac{a}{a+1} + \frac{a}{a+1} = \frac{2a}{a+1}.
\]

3. The constant terms \(1 + 1 + 1 = 3\).

4. The terms with \(-\frac{a}{a^2}\):

\[
-\frac{a}{a^2} - \frac{a}{a^2} = -\frac{2a}{a^2} = -\frac{2}{a}. \quad (\text{assuming } a \neq 0)
\]

5. Finally, the other \(2a\) terms:

\[
2a + 2a = 4a.
\]

Now, combining everything, we have:

\[
\frac{14a}{(a+1)^2} + \frac{2a}{a+1} + 4a + 3 - \frac{2}{a}.
\]

Now the expression is simplified. We can rewrite it compactly:

1. The combined expression is:

\[
\frac{14a}{(a+1)^2} + \frac{2a}{a+1} + 4a + 3 - \frac{2}{a}.
\]

Thus, the final simplified expression is:

\[
\frac{14a}{(a+1)^2} + \frac{2a}{a+1} + 4a + 3 - \frac{2}{a}.
\]