Answer
$$t \sqrt{t^{2}+3}+3 \ln \left|\frac{\sqrt{t^{2}+3}+t}{\sqrt{3}}\right|+C$$
Work Step by Step
Given $$ \sqrt{12+4 t^{2}} d t $$
Let $$ t= \sqrt{3}\tan \theta \ \ \ dt= \sqrt{3}\sec^2 \theta d \theta $$
Then
\begin{aligned}
\int \sqrt{12+4 t^{2}} d t &=\int \sqrt{4\left(3+t^{2}\right)} d t \\
&=2 \int \sqrt{3+t^{2}} d t \\
&=2 \int \sqrt{3+(\sqrt{3} \tan \theta)^{2}} \cdot \sqrt{3} \sec ^{2} \theta d \theta \\
&=2 \int \sqrt{3\left(1+\tan ^{2} \theta\right)^{2}} \cdot \sqrt{3} \sec ^{2} \theta d \theta \\
&=2 \int 3 \sqrt{\sec ^{2} \theta} \cdot \sec ^{2} \theta d \theta \\
&=6 \int \sec \theta \cdot \sec ^{2} \theta d \theta \\
&=6 \int \sec ^{3} \theta d \theta
\end{aligned}
Use
$$ \int \sec ^{m} x d x=\frac{\tan x \sec ^{m-2} x}{m-1}+\frac{m-2}{m-1} \int \sec ^{m-2} x d x$$
Then
\begin{aligned}
\int \sqrt{12+4 t^{2}} d t &=6\left[\frac{\tan \theta \sec ^{3-2} \theta}{3-1}+\frac{3-2}{3-1} \int \sec ^{3-2} \theta d \theta\right]\\
&=6\left[\frac{\tan \theta \sec \theta}{2}+\frac{1}{2} \int \sec \theta d \theta\right]\\
&=6 \cdot \frac{1}{2}[\tan \theta \sec \theta+\ln |\sec \theta+\tan \theta|]+C\\
&=3 \tan \theta \sec \theta+3 \ln |\sec \theta+\tan \theta|+C\\
&=t \sqrt{t^{2}+3}+3 \ln \left|\frac{\sqrt{t^{2}+3}+t}{\sqrt{3}}\right|+C
\end{aligned}
