在 ```\left[...\right]``` 内换行

Question

我想打破 之内的界限\left[...\right]。

\documentclass[12pt] {article}
\usepackage{epsf}
\usepackage{amsmath}
\usepackage{amssymb}

\begin{document}
\begin{align*}
            \frac{\partial U}{\partial k} &=x^k\log x
            \left[1-\frac{1}{(k+2)^2}x^{2}+\frac{1}{(k+2)^2(k+4)^2}x^{4}-\frac{1}{(k+2)^2(k+4)^2(k+6)^2}x^6+....\right]+x^k\\
            & \left[\frac{2}{(k+2)^3}x^{2}+\frac{1}{(k+2)^2(k+4)^2}\left\{-\frac{2}{k+2}-\frac{2}{k+4}\right\}x^{4}
            -\frac{1}{(k+2)^2(k+4)^2(k+6)^2}\left\{-\frac{2}{k+2}\right.\\
            &\left.-\frac{2}{k+4}-\frac{2}{k+6}\right\}x^6+....\right]
    \end{align*}
\end{document}

Answer 1

---

\documentclass[12pt]{article}
\usepackage{amsmath,geometry}
\usepackage{amssymb}

\begin{document}
\begin{align*}
            \frac{\partial U}{\partial k} &=x^k\log x
            \Biggl[1-\frac{1}{(k+2)^2}x^{2}+\frac{1}{(k+2)^2(k+4)^2}x^{4}-\frac{1}{(k+2)^2(k+4)^2(k+6)^2}x^6+\dotsb\Biggr]+x^k\\
            & \Biggl[\frac{2}{(k+2)^3}x^{2}+\frac{1}{(k+2)^2(k+4)^2}\left\{-\frac{2}{k+2}-\frac{2}{k+4}\right\}x^{4}\\
            &-\frac{1}{(k+2)^2(k+4)^2(k+6)^2}\left\{-\frac{2}{k+2}-\frac{2}{k+4}-\frac{2}{k+6}\right\}x^6+\dotsb\Biggr]
            \end{align*}
\end{document}