Integral with fraction and square root produces wrong vertical alignment

Question

This code produces an ankward vertical alignment which involves the integral symbol and the fraction with the square root at the denominator. Why? How can I fix this?

\documentclass[a4paper]{article}

\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{mathtools}

\begin{document}
 \[
 \int_{x_0}^{x(t)} \! \! \! \frac{1}{\sqrt{\frac{2}{m}\bigl
 (E_0-V(x)\bigr)}} \, dx = t-t_0
 \]
\end{document}

Answer 1

That's how it's supposed to be typeset.

Growing the integral symbol is out of the question, in my opinion. Moving it down is similarly disgraceful.

\documentclass[a4paper]{article}

\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{mathtools}

\begin{document}

\noindent
No square root
 \[
 \int_{x_0}^{x(t)}
   \biggl(\frac{2}{m}\bigl(E_0-V(x)\bigr)\biggr)^{\!-1/2}\, dx = t-t_0
 \]
Smashed square root
 \[
 \int_{x_0}^{x(t)}
   \frac{1}{\sqrt{\smash[b]{\frac{2}{m}}(E_0-V(x))}} \, dx = t-t_0
 \]
Normal
 \[
 \int_{x_0}^{x(t)}
   \frac{1}{\sqrt{\frac{2}{m}\bigl(E_0-V(x)\bigr)}} \, dx = t-t_0
 \]
\end{document}

Answer 2

I probably would do

\documentclass[a4paper]{article}
\usepackage{amsmath}

\begin{document}
 \[
 \int\limits_{x_0}^{x(t)}  \frac{1}{\sqrt{\frac{2}{m}\bigl
 (E_0-V(\xi)\bigr)}} \, \mathrm{d}\xi = t-t_0
 \]
\end{document}

to avoid confusion between the upper limit of the integral, x(t), and the integration variable as well between some quantity d (like a diameter, say) and the differential d. \int\limits makes the integral symbol look higher.

Answer 3

As marmot suggested, I tried using the scalerel package and I like the outcome. It's not perfect, but I think it's acceptable. The integral symbol is finally aligned with the bottom of the fraction.

\documentclass[a4paper]{article}

\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{mathtools}

\usepackage{scalerel}

\begin{document}
 \def\x{\frac{1}{\sqrt{\frac{2}{m}\bigl(E_0-V
 (\xi)\bigr)}}}
 \[
 {\scalerel*[2.2ex]{\int}{\x}}_{\!\!\!x_0}^{\,x(t)}\x\,\,
 d\xi \, = \, t-t_0
 \]
\end{document}