Answer
$$\int \frac{d t}{t \sqrt{\left(3+ t^{2}\right)}}=-\frac{1}{\sqrt{3}} \operatorname{csch}^{-1}\left|\frac{t}{\sqrt{3}}\right|+C $$
Work Step by Step
Given $$\int \frac{d t}{t \sqrt{\left(3+ t^{2}\right)}}$$
So, we have
\begin{aligned}
I&=\int \frac{d t}{t \sqrt{\left(3+ t^{2}\right)}}=\\
&\int \frac{d t}{t \sqrt{3\left(1+\frac{t^{2}}{3}\right)}}\\
&=\int \frac{d t}{\sqrt{3 }t \sqrt{1+\frac{t^{2}}{3}}}
\end{aligned}
Let
$$ u=\frac{t}{\sqrt{3}},\Rightarrow d u=\frac{d t}{\sqrt{3}}$$
So, we get
\begin{aligned}
I& =\int \frac{\sqrt{3} d u}{\sqrt{3}(\sqrt{3} u) \sqrt{1+u^{2}}}\\
&=\frac{1}{\sqrt{3}} \int \frac{d u}{u \sqrt{1+u^{2}}}\\
&=-\frac{1}{\sqrt{3}} \operatorname{csch}^{-1}|u|+C\\
&=-\frac{1}{\sqrt{3}} \operatorname{csch}^{-1}\left|\frac{t}{\sqrt{3}}\right|+C
\end{aligned}
