3

In my attached code I'm trying to cite two outputs from Wolfram Alpha and I cannot get the URL to show up at the end of my document. Only the author and title will render. I'm not sure what to do.

Here is my code:

\documentclass[11pt]{amsart}
\usepackage{amssymb,latexsym,url,graphicx}
\topmargin 0in
\usepackage{fullpage}
\usepackage{array}
\usepackage{filecontents}
\usepackage{url} 
\theoremstyle{definition}
\newtheorem{definition}{Definition}
\newtheorem{solution}{Solution}
\newtheorem{problem}{Problem}
\newtheorem{corollary}{Corollary}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{example}[theorem]{Example}
\begin{document}
\title{CALCULUS II: OPTIONAL PROJECT}
\maketitle
\section{Calculus techniques from Schwarzschild (Question 4)}
Now we find the exact distance between the shells using integration (instead of just estimating it). This works better, because it uses the change in relativity over the one meter distance. Thus, using Wolfram Alpha\cite{Wolfram1}:
\begin{align*}
\Delta r_{\text{shell}}&=\int_{r_1}^{r_2}\frac{r^{\frac{1}{2}}}{(r-2M)^{\frac{1}{2}}}dr \\
&=\int_{30001}^{30002}\frac{r^{\frac{1}{2}}}{\left[(r-2(14933))\right]^{\frac{1}{2}}}dr \\
&\approx 14.880 \text{ meters}\\
\end{align*}
Note: One can notice how the above answer is between our two previous estimates. 
\section{How many shells? (Question 5)}
One can assume that the rods that the robot drones use to measure 1 meter to the next shell are infinitely rigid (that is, the rods are strong enough that the relativity from gravity doesn't effect their length - so their directly measured length\footnote{Directly Measured distance = The "Real" distance (aka. the distance after taking into account relativity)} is always 1 meter). Thus, using Wolfram Alpha\cite{Wolfram2}: 
\begin{align*}
\Delta r_{\text{shell}}&=\int_{r_1}^{r_2}\frac{r^{\frac{1}{2}}}{(r-2M)^{\frac{1}{2}}}dr \\
&=\int_{30000}^{31000}\frac{r^{\frac{1}{2}}}{\left[(r-2(14933))\right]^{\frac{1}{2}}}dr \\
&\approx 7708.5 \text{ meters}\\
\end{align*}
\newpage
\bibliography{project2} 
\bibliographystyle{plain}
\end{document}

Here is my project2.bib code:

@online{Wolfram1, 
author = {Wolfram Alpha},
title  = {Wolfram Alpha},
date   = {2016-12},
url    = {https://www.wolframalpha.com/input/?i=\%5Cint_\%7B30000\%7D\%5E\%7B31000\%7D\%5Cfrac\%7Br\%5E\%7B\%5Cfrac\%7B1\%7D\%7B2\%7D\%7D\%7D\%7B\%5Cleft\%5B(r-2(14933))\%5Cright\%5D\%5E\%7B\%5Cfrac\%7B1\%7D\%7B2\%7D\%7D\%7Ddr}
}
@online{Wolfram2, 
author = {Wolfram Alpha},
title  = {Wolfram Alpha},
date   = {2016-12},
url    = {https://www.wolframalpha.com/input/?i=\%5Cint_\%7B30001\%7D\%5E\%7B30002\%7D\%5Cfrac\%7Br\%5E\%7B\%5Cfrac\%7B1\%7D\%7B2\%7D\%7D\%7D\%7B\%5Cleft\%5B(r-2(14933))\%5Cright\%5D\%5E\%7B\%5Cfrac\%7B1\%7D\%7B2\%7D\%7D\%7Ddr}
}
1

Just give biblatex a chance (important code changings marked with <==============):

\RequirePackage{filecontents}
\begin{filecontents*}{\jobname.bib}
@misc{Wolfram1, 
  author = {Wolfram Alpha},
  title  = {Wolfram Alpha},
  date   = {2016-12},
  url    = {https://www.wolframalpha.com/input/?i=\%5Cint_\%7B30000\%7D\%5E\%7B31000\%7D\%5Cfrac\%7Br\%5E\%7B\%5Cfrac\%7B1\%7D\%7B2\%7D\%7D\%7D\%7B\%5Cleft\%5B(r-2(14933))\%5Cright\%5D\%5E\%7B\%5Cfrac\%7B1\%7D\%7B2\%7D\%7D\%7Ddr}
}
@misc{Wolfram2, 
  author = {Wolfram Alpha},
  title  = {Wolfram Alpha},
  date   = {2016-12},
  url    = {https://www.wolframalpha.com/input/?i=\%5Cint_\%7B30001\%7D\%5E\%7B30002\%7D\%5Cfrac\%7Br\%5E\%7B\%5Cfrac\%7B1\%7D\%7B2\%7D\%7D\%7D\%7B\%5Cleft\%5B(r-2(14933))\%5Cright\%5D\%5E\%7B\%5Cfrac\%7B1\%7D\%7B2\%7D\%7D\%7Ddr}
}
\end{filecontents*}


\documentclass[11pt]{amsart}

\usepackage{amssymb,latexsym,graphicx}
%\topmargin 0in
\usepackage{fullpage}
\usepackage{array}

\usepackage{url} 
\usepackage{biblatex} % <===============================================
\addbibresource{\jobname.bib} % <=======================================

\theoremstyle{definition}
\newtheorem{definition}{Definition}
\newtheorem{solution}{Solution}
\newtheorem{problem}{Problem}
\newtheorem{corollary}{Corollary}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{example}[theorem]{Example}


\begin{document}
\title{CALCULUS II: OPTIONAL PROJECT}
\maketitle
\section{Calculus techniques from Schwarzschild (Question 4)}
Now we find the exact distance between the shells using integration (instead of just estimating it). This works better, because it uses the change in relativity over the one meter distance. Thus, using Wolfram Alpha\cite{Wolfram1}:
\begin{align*}
\Delta r_{\text{shell}}&=\int_{r_1}^{r_2}\frac{r^{\frac{1}{2}}}{(r-2M)^{\frac{1}{2}}}dr \\
&=\int_{30001}^{30002}\frac{r^{\frac{1}{2}}}{\left[(r-2(14933))\right]^{\frac{1}{2}}}dr \\
&\approx 14.880 \text{ meters}\\
\end{align*}
Note: One can notice how the above answer is between our two previous estimates. 
\section{How many shells? (Question 5)}
One can assume that the rods that the robot drones use to measure 1 meter to the next shell are infinitely rigid (that is, the rods are strong enough that the relativity from gravity doesn't effect their length - so their directly measured length\footnote{Directly Measured distance = The "Real" distance (aka. the distance after taking into account relativity)} is always 1 meter). Thus, using Wolfram Alpha\cite{Wolfram2}: 
\begin{align*}
\Delta r_{\text{shell}}&=\int_{r_1}^{r_2}\frac{r^{\frac{1}{2}}}{(r-2M)^{\frac{1}{2}}}dr \\
&=\int_{30000}^{31000}\frac{r^{\frac{1}{2}}}{\left[(r-2(14933))\right]^{\frac{1}{2}}}dr \\
&\approx 7708.5 \text{ meters}\\
\end{align*}
\newpage
%\bibliography{\jobname} 
%\bibliographystyle{plain}
\printbibliography % <==================================================
\end{document}

with the result:

enter image description here

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