Given,
$$
\begin{aligned}
& \int \frac{x^2}{2 x^2+6 \sqrt{2} x+9} d x \\
& \int \frac{\frac{1}{2}\left(2 x^2+6 \sqrt{2} x+9\right)-\left(3 \sqrt{2} x+\frac{9}{2}\right)}{2 x^2+6 \sqrt{2} x+9} d x \\
& {\left[\therefore x^2=\left(2 x^2+6 \sqrt{2} x+9\right) \frac{1}{2}-\left(3 \sqrt{2} x+\frac{9}{2}\right)\right]} \\
& \int\left(\frac{1}{2}-\frac{\left(3 \sqrt{2} x+\frac{9}{2}\right)}{2 x^2+6 \sqrt{2} x+9}\right) d x \\
& \int \frac{1}{2} d x-\int \frac{3 \sqrt{2} x+\frac{9}{2}}{2 x^2+6 \sqrt{2} x+9} d x \\
&
\end{aligned}
$$


$3 \sqrt{2} x+\frac{9}{2}=A \frac{d}{d x}\left(2 x^2+6 \sqrt{2} x+9\right)+B$

On comparison, we get
$$
\begin{aligned}
& 4 A=3 \sqrt{2} \\
& 6 \sqrt{2} A+B=\frac{9}{2} \\
& A=\frac{3 \sqrt{2}}{4} \\
& 6 \sqrt{2} \times \frac{3 \sqrt{2}}{4}+B=\frac{9}{2} \\
& B=-\frac{9}{2}
\end{aligned}
$$

Substituting values of $A, B$ Eq. (ii)

Substituting Eq. (iii) in Eq. (i)
$$
\begin{gathered}
\frac{x}{2}-\int \frac{\left(\frac{3 \sqrt{2}}{4}(4 x+6 \sqrt{2})-\frac{9}{2}\right) d x}{2 x^2+6 \sqrt{2} x+9} \\
\frac{x}{2}-\frac{3 \sqrt{2}}{4} \int \frac{4 x+6 \sqrt{2} d x}{2 x^2+6 \sqrt{2} x+9}+\frac{9}{2} \int \frac{1}{2 x^2+6 \sqrt{2} x+9} d x \\
\frac{x}{2}-\frac{3 \sqrt{2}}{4} \log \left|2 x^2+6 \sqrt{2} x+9\right|+\frac{9}{2} \int \frac{1}{(\sqrt{2} x+3)^2} d x \\
\\
{\left[\because \int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|\right]}
\end{gathered}
$$
$$
\begin{aligned}
& \frac{x}{2}-\frac{3 \sqrt{2}}{4} \log \left|(\sqrt{2} x+3)^2\right|+\frac{9}{2} \frac{(\sqrt{2} x+3)^{-2+1}}{-2+1} \times \frac{1}{\sqrt{2}}+c^{\prime} \\
& \frac{x}{2}-\frac{3 \sqrt{2}}{4} \times 2 \log |\sqrt{2} x+3|+\frac{9}{2} \frac{(\sqrt{2} x+3)^{-1}}{-1} \times \frac{1}{\sqrt{2}}+c^{\prime} \\
& \frac{x}{2}-\frac{3 \sqrt{2}}{2} \log |\sqrt{2} x+3|-\frac{9}{2 \sqrt{2}} \times \frac{1}{\sqrt{2} x+3}+c^{\prime} \\
& \frac{1}{2 \sqrt{2}}\left[\sqrt{2} x-6 \log |\sqrt{2} x+3|-\frac{9}{\sqrt{2} x+3}\right]+c^{\prime} \\
& =\frac{1}{2 \sqrt{2}}\left[(\sqrt{2} x+3)-6 \log |\sqrt{2} x+3|-\frac{9}{\sqrt{2} x+3}\right]+c
\end{aligned}
$$

Hence, option (a) is correct.