I have two expressions, they need to be the same and equal but I can`t see where I am making mistake:
$\int (p_0 dp_1+p_1 dp_0)=\int d(p_1p_0)$
On the other side, when I integrate both sides I have result:
$p_0p_1+p_1p_0+C_1=p_1p_0+C_2$
What I have missed here?
$\int (p_0 dp_1+p_1 dp_0)=\int d(p_1p_0)\qquad$ : This is correct.
$\int d(p_1p_0)=p_1p_0+C_2\qquad$ : This is correct.
$\int (p_0 dp_1+p_1 dp_0)=p_0p_1+p_1p_0+C_1\qquad$ : This is false.
Because $\quad\int p_0 dp_1 \neq p_0p_1+c_1\quad$ and $\quad\int p_1 dp_0 \neq p_1p_0+c_2$
One cannot integrate separately $\int p_0 dp_1$ and separately $\int p_0 dp_1$. One have to integrate them together, that is $\int (p_0 dp_1+p_1 dp_0)$ in writing : $$\int (p_0 dp_1+p_1 dp_0)=\int d(p_1p_0)=p_1p_0+C_2$$ The term on the right gives the result for the whole term on the left.
NOTE :
One can integrate $\int f(x)dx$ because the function $f(x)$ is function of the same variable than the variable of integration $x$.
One cannot integrate $\int f(x)dt$ because the function $f(x)$ is function of $x$ but not of $t$. More exactly, if $f(x)$ is function of $x$ only, not function of $x$ and $t$, then $\int f(x)dt=f(x)\int dt=f(x)(t+c)$.
One can integrate $\int f(x(t))dt$ because the function of function $f(x(t))$ is function of the same variable than the variable of integration $t$.
In the case $\int p_0dp_1$ it is not specified that $p_0$ is function of $p_1$. So, this is the same case as $\int f(x)dt$ above.