Answer
$$x-\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right)+C$$
Work Step by Step
Given $$\int \frac{x^{2} d x}{x^{2}+3}$$
Since
\begin{align*}
\int \frac{x^{2} d x}{x^{2}+3}&=\int \frac{(x^{2}+1-1) d x}{x^{2}+3}\\
&=\int \left( 1-\frac{1}{x^2+3}\right)dx\\
&=x-\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right)+C
\end{align*}