Since $r^n+a_1r^{n-1}+...+a_n=0$ is an equation of integral dependence of $r$ over an ideal $I$, by definition, $a_i\in I^i$. So
$a_1\in I$ is generated by one element of $I$. till now nothing goes wrong if we consider the ideal $(a_1)$ instead of $I$.
$a_2\in I^2$; let $a_2 $ is sum of products like $t_{k1} t_{l2}$ (t_k1 ,t_l2 are elements of I). so $a_2$ is generated by finite elements of $I$. till now, nothing goes wrong if we consider the ideal $(a_1,\{t_{k1}\}_k,\{t_{l2}\}_l)$ instead of $I$
...
write the same argument for $a_i$. and consider the ideal generated by these finite elements

Since $r^n+a_1r^{n-1}+...+a_n=0$ is an equation of integral dependence of $r$ over an ideal $I$, by definition, $a_i\in I^i$. So $\displaystyle a_i=\sum_{k=1}^{k=n(i)}a_{k1}^{(i)}\ldots a_{ki}^{(i)}$, with $a_{kj}^{(i)}\in I$. Take for $J$ the ideal generated by the $a_{kj}^{(i)}$.

Since $r^n+a_1r^{n-1}+...+a_n=0$ is an equation of integral dependence of $r$ over an ideal $I$, by definition, $a_i\in I^i$. So
$a_1\in I$ is generated by one element of $I$. till now nothing goes wrong if we consider the ideal $(a_1)$ instead of $I$.
$a_2\in I^2$; let $a_2 = t_1 t_2$ (t_1 ,t_2 are elements of I). so $a_2$ is generated by two elements of $I$. till now, nothing goes wrong if we consider the ideal $(a_1,t_1,t_2)$ instead of $I$
...
write the same argument for $a_i$. and consider the ideal generated by these finite elements

Since $r^n+a_1r^{n-1}+...+a_n=0$ is an equation of integral dependence of $r$ over an ideal $I$, by definition, $a_i\in I^i$. So
$a_1\in I$ is generated by one element of $I$. till now nothing goes wrong if we consider the ideal $(a_1)$ instead of $I$.
$a_2\in I^2$; let $a_2 $ is sum of products like $t_{k1} t_{l2}$ (t_k1 ,t_l2 are elements of I). so $a_2$ is generated by finite elements of $I$. till now, nothing goes wrong if we consider the ideal $(a_1,\{t_{k1}\}_k,\{t_{l2}\}_l)$ instead of $I$
...
write the same argument for $a_i$. and consider the ideal generated by these finite elements