split of equation

Question

How can I please get this equation to center align?

\documentclass[12pt, a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
\usepackage[english]{babel}
\usepackage{physics}
\usepackage{mathtools, nccmath}
\usepackage{calc}
\usepackage{float}
\usepackage{fancyhdr}
\usepackage[makeroom]{cancel}
\usepackage{mdframed}
\usepackage[showframe]{geometry}

\begin{document}
\mbox{}
\begin{fleqn}
\begin{equation}\label{in1}
\begin{aligned}[t]
 & \textcolor{blue}{\int_0^{r_1} r_2^2 e^{-\frac{2Zr_2}{a}} \mathrm{d}r_2} =\begin{vmatrix}
u = r_2^2 &u' = 2r_2 \\
v'=e^{-\frac{2Zr_2}{a}} &v = -\frac{a}{2Z} e^{-\frac{2Zr_2}{a}}
\end{vmatrix} 
= \left[\frac{r_2^2 e^{-\frac{2zr_1}{a}}}{-\frac{2Z}{a}}  \right]_0^{r_1} + \int_0^{r_1} r_2 \frac{a}{Z} e^{-\frac{2Zr_2}{a}} \mathrm{d}r_2 } = \\
 & = \left[\frac{r_2^2 e^{-\frac{2zr_1}{a}}}{-\frac{2Z}{a}}  \right]_0^{r_1} + \left[-\frac{a^2}{Z}r_2 \frac{e^{-\frac{2Zr_2}{a}}}{2Z} \right]_0^{r_1} + \int_0^{r_1} \frac{ a^2}{2Z^2} e^{-\frac{2Zr_2}{a}}  \mathrm{d}r_2 = \\
 & = \left[\frac{r_2^2 e^{-\frac{2zr_1}{a}}}{-\frac{2Z}{a}}  \right]_0^{r_1} + \left[-\frac{a^2}{Z}r_2 \frac{e^{-\frac{2Zr_2}{a}}}{2Z} \right]_0^{r_1} + \left[-\frac{a^3}{4Z^3}e^{-\frac{2Zr_2}{a}} \right]_0^{r_1} = \\
 & = -\frac{ar_1^2e^{-\frac{2Zr_1}{a}}}{2Z} - \frac{a^2}{2Z^2}r_1e^{-\frac{2Zr_1}{a}}-\frac{a^3}{4Z^3} \left(e^{-\frac{2Zr_1}{a}}-1 \right)
\end{aligned}
\end{equation}
\end{fleqn}



\mbox{}
\begin{fleqn}
\begin{equation}\label{in2}
\begin{aligned}[t]
\textcolor{red}{\int_{r_1}^{\infty} r_2 e^{-\frac{2Zr_2}{a}} \mathrm{d}r_2 } = \begin{vmatrix}
u = r_2 &u' = 1 \\
v'=e^{-\frac{2Zr_2}{a}} &v = -\frac{a}{2Z} e^{-\frac{2Zr_2}{a}}
\end{vmatrix} =  \\
& = \left[-\frac{ar_2}{2Z}e^{-\frac{2Zr_2}{a}} \right]_{r_1}^{\infty} + \int_{r_1}^{\infty} \frac{a}{2Z}e^{-\frac{2Zr_2}{a}}\mathrm{d}r_2 } = \\
& = \left[-\frac{ar_2}{2Z}e^{-\frac{2Zr_2}{a}} \right]_{r_1}^{\infty} + \left[-\frac{ar_2}{4Z^2}e^{-\frac{2Zr_2}{a}} \right]_{r_1}^{\infty} = \\ 
& = \frac{ar_1}{2Z}e^{-\frac{2Zr_1}{a}}+\frac{a^2}{4Z^2}e^{-\frac{2Zr_1}{a}}
\end{aligned}
\end{equation}
\end{fleqn}
\end{document}

Answer 1

You just need to play around a little with the alignment operator &. You achieve alignment across several equation blocks via this answer and adjust the alignment position by changing the values in the command definition. It is not the prettiest solution, but simple and does what it needs to do.

I have also taken the liberty of cleaning up your code and use the align environment instead of three nested environments.

\documentclass[12pt, a4paper]{article}
\usepackage{amsmath}
\usepackage{xcolor}
\newcommand{\fakealign}{%
   \mbox{\hspace{2cm}} & \mbox{\hspace{12cm}} \nonumber\\}
\begin{document}
\begin{align}
\fakealign
\textcolor{blue}{\int_0^{r_1} r_2^2 e^{-\frac{2Zr_2}{a}} \mathrm{d}r_2} &=\begin{vmatrix}
u = r_2^2 &u' = 2r_2 \\
v'=e^{-\frac{2Zr_2}{a}} &v = -\frac{a}{2Z} e^{-\frac{2Zr_2}{a}}
\end{vmatrix} 
= \left[\frac{r_2^2 e^{-\frac{2zr_1}{a}}}{-\frac{2Z}{a}}  \right]_0^{r_1} + \int_0^{r_1} r_2 \frac{a}{Z} e^{-\frac{2Zr_2}{a}} \mathrm{d}r_2 = \\
 & = \left[\frac{r_2^2 e^{-\frac{2zr_1}{a}}}{-\frac{2Z}{a}}  \right]_0^{r_1} + \left[-\frac{a^2}{Z}r_2 \frac{e^{-\frac{2Zr_2}{a}}}{2Z} \right]_0^{r_1} + \int_0^{r_1} \frac{ a^2}{2Z^2} e^{-\frac{2Zr_2}{a}}  \mathrm{d}r_2 = \\
 & = \left[\frac{r_2^2 e^{-\frac{2zr_1}{a}}}{-\frac{2Z}{a}}  \right]_0^{r_1} + \left[-\frac{a^2}{Z}r_2 \frac{e^{-\frac{2Zr_2}{a}}}{2Z} \right]_0^{r_1} + \left[-\frac{a^3}{4Z^3}e^{-\frac{2Zr_2}{a}} \right]_0^{r_1} = \\
 & = -\frac{ar_1^2e^{-\frac{2Zr_1}{a}}}{2Z} - \frac{a^2}{2Z^2}r_1e^{-\frac{2Zr_1}{a}}-\frac{a^3}{4Z^3} \left(e^{-\frac{2Zr_1}{a}}-1 \right)
\end{align}

\begin{align}
\fakealign
\textcolor{red}{\int_{r_1}^{\infty} r_2 e^{-\frac{2Zr_2}{a}} \mathrm{d}r_2 } &= \begin{vmatrix}
u = r_2 &u' = 1 \\
v'=e^{-\frac{2Zr_2}{a}} &v = -\frac{a}{2Z} e^{-\frac{2Zr_2}{a}}
\end{vmatrix} =  \\
& = \left[-\frac{ar_2}{2Z}e^{-\frac{2Zr_2}{a}} \right]_{r_1}^{\infty} + \int_{r_1}^{\infty} \frac{a}{2Z}e^{-\frac{2Zr_2}{a}}\mathrm{d}r_2 = \\
& = \left[-\frac{ar_2}{2Z}e^{-\frac{2Zr_2}{a}} \right]_{r_1}^{\infty} + \left[-\frac{ar_2}{4Z^2}e^{-\frac{2Zr_2}{a}} \right]_{r_1}^{\infty} = \\ 
& = \frac{ar_1}{2Z}e^{-\frac{2Zr_1}{a}}+\frac{a^2}{4Z^2}e^{-\frac{2Zr_1}{a}}
\end{align}
\end{document}

Answer 2

You simply have to remove the fleqn environment: it makes equations start at the leftmargin (like the fleqn document class option, but only within the environment). So hopefully the fllowing code produces what you want:

\documentclass[12pt, a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
\usepackage[english]{babel}
\usepackage{physics}
\usepackage{mathtools, nccmath}
\usepackage{calc}
\usepackage{float}
\usepackage{fancyhdr}
\usepackage[makeroom]{cancel}
\usepackage{mdframed}
\usepackage[showframe]{geometry}

\begin{document}

\begin{equation}\label{in1}
\begin{aligned}[t]
\textcolor{blue}{\int_0^{r_1} r_2^2 e^{-\frac{2Zr_2}{a}} \mathrm{d}r_2} &=\begin{vmatrix}
u = r_2^2 &u' = 2r_2 \\
v'=e^{-\frac{2Zr_2}{a}} &v = -\frac{a}{2Z} e^{-\frac{2Zr_2}{a}}
\end{vmatrix} \\
 & = \left[\frac{r_2^2 e^{-\frac{2zr_1}{a}}}{-\frac{2Z}{a}} \right]_0^{r_1} + \int_0^{r_1} r_2 \frac{a}{Z} e^{-\frac{2Zr_2}{a}}
\mathrm{d}r_2 =
\\
 & = \left[\frac{r_2^2 e^{-\frac{2zr_1}{a}}}{-\frac{2Z}{a}} \right]_0^{r_1} + \left[-\frac{a^2}{Z}r_2 \frac{e^{-\frac{2Zr_2}{a}}}{2Z} \right]_0^{r_1} + \int_0^{r_1} \frac{ a^2}{2Z^2} e^{-\frac{2Zr_2}{a}} \mathrm{d}r_2 = \\
 & = \left[\frac{r_2^2 e^{-\frac{2zr_1}{a}}}{-\frac{2Z}{a}} \right]_0^{r_1} + \left[-\frac{a^2}{Z}r_2 \frac{e^{-\frac{2Zr_2}{a}}}{2Z} \right]_0^{r_1} + \left[-\frac{a^3}{4Z^3}e^{-\frac{2Zr_2}{a}} \right]_0^{r_1} = \\
 & = -\frac{ar_1^2e^{-\frac{2Zr_1}{a}}}{2Z} - \frac{a^2}{2Z^2}r_1e^{-\frac{2Zr_1}{a}}-\frac{a^3}{4Z^3} \left(e^{-\frac{2Zr_1}{a}}-1 \right)
\end{aligned}
\end{equation}
\mbox{}

\begin{equation}\label{in2}
\begin{aligned}[t]
\textcolor{red}{\int_{r_1}^{\infty} r_2 e^{-\frac{2Zr_2}{a}} \mathrm{d}r_2 } & = \begin{vmatrix}
u = r_2 &u' = 1 \\
v'=e^{-\frac{2Zr_2}{a}} &v = -\frac{a}{2Z} e^{-\frac{2Zr_2}{a}}
\end{vmatrix} = \\
& = \left[-\frac{ar_2}{2Z}e^{-\frac{2Zr_2}{a}} \right]_{r_1}^{\infty} + \int_{r_1}^{\infty} \frac{a}{2Z}e^{-\frac{2Zr_2}{a}}\mathrm{d}r_2 = \\
& = \left[-\frac{ar_2}{2Z}e^{-\frac{2Zr_2}{a}} \right]_{r_1}^{\infty} + \left[-\frac{ar_2}{4Z^2}e^{-\frac{2Zr_2}{a}} \right]_{r_1}^{\infty} = \\
& = \frac{ar_1}{2Z}e^{-\frac{2Zr_1}{a}}+\frac{a^2}{4Z^2}e^{-\frac{2Zr_1}{a}}
\end{aligned}
\end{equation}

\end{document}