Answer
$x=3 \tan ^{-1}(\sqrt{3}t)-\sqrt{3} \tan ^{-1}(t)-\pi
$
Work Step by Step
$$ \left(3 t^{4}+4 t^{2}+1\right) \frac{d x}{d t}=2 \sqrt{3}, \quad x(1)=-\pi \frac{\sqrt{3}}{4}$$
$$
\frac{d x}{d t}=\frac{2 \sqrt{3}}{\left(3 t^{4}+4 t^{2}+1\right)} \rightarrow d x=\frac{2 \sqrt{3}}{\left(3 t^{4}+4 t^{2}+1\right)} d t
$$
$$d x=\frac{2 \sqrt{3}}{\left(3 u^{2}+4 u+1\right)} d t \quad$$
$$
\begin{aligned}
&\text {"We do a substitution we say $u$ $=$ $t^{2}$, then we factor it"}\\
\end{aligned}
$$
$$d x=\frac{2 \sqrt{3}}{(3 u+1)(u+1)}d t$$
$$
\begin{aligned}
&\text {"Now we replace $u$ for $t^{2}$ "}\\
\end{aligned}
$$
$$d x=\frac{2 \sqrt{3}}{\left(3 t^{2}+1\right)\left(t^{2}+1\right)} d t$$
$$
\begin{aligned}
&\text {"Now we continue integrating on both sides"}\\
\end{aligned}
$$
$$
\begin{aligned}
&\int d x=\int \frac{2 \sqrt{3}}{3\left(t^{2}+\frac{1}{3}\right)\left(t^{2}+1\right)} d t \\
&x+C=\frac{2 \sqrt{3}}{3} \int \frac{1}{\left(t^{2}+\frac{1}{3}\right)\left(t^{2}+1\right)} d t \\
&\text {"Now we do partial fraction to do the integral"}\\
&\left[\frac{1}{\left(t^{2}+\frac{1}{3}\right)\left(t^{2}+1\right)}=\frac{A t+B}{t^{2}+\frac{1}{3}}+\frac{C t+D}{t^{2}+1}\right]\left(t^{2}+\frac{1}{3}\right)\left(t^{2}+1\right) \\
&1=(A t+B)\left(t^{2}+1\right)+(C t+D)\left(t^{2}+\frac{1}{3}\right) \\
&1=A t^{3}+A t+B t^{2}+B+C t^{3}+\frac{1}{3} C t+D t^{2}+\frac{1}{3} \\
&1=[A+C] t^{3}+[B+D] t^{2}+[A+\frac{1}{3} C] t+[B+\frac{1}{3} D]
\end{aligned}
$$
$$
\begin{aligned}
&\text {"Now we set our set of equations"}\\
\end{aligned}
$$
\begin{align}
\begin{aligned}
& A+C=0 \\
& B+D=0 \\
& A+\frac{1}{3}C=0 \\
& B+ \frac{1}{3}D=1 \\
\end{aligned}
\end{align}
$$
\begin{aligned}
&\quad\quad\quad A=0 ;\quad B=\frac{3}{2} ;\quad C=0 ;\quad D=-\frac{3}{2} \\
&\text {"When we have the value of each letter we replace it on our partial fraction"}\\\\
&x+C=\frac{2 \sqrt{3}}{3} \int \frac{3 / 2}{t^{2}+1 / 3}d t+\frac{-3 / 2}{t^{2}+1} d t \\
&x+C=\frac{2 \sqrt{3}}{3}\left[\frac{3}{2} \int \frac{1}{t^{2}+1 / 3}d t -\frac{3}{2} \int \frac{1}{t^{2}+1} d t\right] \\
&x+C=\frac{2 \sqrt{3}}{3}\left(\frac{3}{2}\right)\left[\int \frac{1}{t^{2}+1 / 3} d t-\int \frac{1}{t^{2}+1} d t\right] \\
&x+C=\frac{2 \sqrt{3}}{3} \cdot \frac{3}{2}\left[\int \frac{1}{t^{2}+1 / 3} d t-\int \frac{1}{t^{2}+1} d t\right] \\
&\text {"Now we solve both integrals"}\\
&x+C=\sqrt{3}\left[\int \frac{1}{t^{2}+1 / 3} d t-\int \frac{1}{t^{2}+1} d t\right]\\
&\text {"Now let's remember the formula of how to integrate this"}\\
\end{aligned}
$$
$$
\int \frac{d u}{u^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{u}{a}\right)+C
$$
"In both integrals the " $u$ " will be " $t$ " but in the first $a^{2}=1 / 3$ and in
the second $a^{2}=1$ so the first $a=\sqrt{1 / 3}$ and in the second $a=\sqrt{1}=1$."
$$
\begin{aligned}
& x+C=\sqrt{3}\left[\frac{1}{(\sqrt{3} / 3)} \tan ^{-1}\left(\frac{t}{\sqrt{3} / 3}\right)-\frac{1}{1} \tan ^{-1}\left(\frac{t}{1}\right)+C\right] \\
& x+C=\sqrt{3}\left[\frac{3}{\sqrt{3}} \tan ^{-1}\left(\frac{3 t}{\sqrt{3}}\right)-\tan ^{-1}(t)+C\right]
\end{aligned}
$$
"$C$ is a constant so we can leave just one because we don't know the value "
$$
\begin{aligned}
& x+C=3 \cdot \tan ^{-1}(\sqrt{3} t)-3 \sqrt{3} \tan ^{-1}(t) \\
& x=3 \tan ^{-1}(\sqrt{3} t)-\sqrt{3} \tan ^{-1}(t)+C
\end{aligned}
$$
"Let's remember $x$ depends on $t$ "
$$
\begin{aligned}
& x(t)=3 \tan ^{-1}(\sqrt{3} t)-\sqrt{3} \tan ^{-1}(t)+C \\
& x(1)=-\frac{\sqrt{3} \pi}{4}
\end{aligned}
$$
"Now let's evaluate in $t=1$ and that will be equal to $-\frac{\sqrt{3}\pi}{4} $"
$$
\begin{aligned}
& 3 \tan ^{-1}(\sqrt{3}(1))-\sqrt{3} \tan ^{-1}(1)+C=-\frac{\sqrt{3}\pi}{4} \\
& 3\left(\frac{\pi}{3}\right)-\sqrt{3}\left(\frac{\pi}{4}\right)+C=-\frac{\sqrt{3}\pi}{4} \\
& \pi-\frac{\sqrt{3}\pi}{4} +C=-\frac{\sqrt{3}\pi}{4} \\
& C=-\frac{\sqrt{3} \pi}{4}+\frac{\sqrt{3} \pi}{4}-\pi \\
& C=-\pi \\
\end{aligned}
$$
"Now we replace C on the answer of x"
$$
\begin{aligned}
& x=3 \tan ^{-1}(\sqrt{3}t)-\sqrt{3} \tan ^{-1}(t)-\pi
\end{aligned}
$$
