Answer
$$\frac{1}{2 \sqrt{6}} \ln |\sqrt{2} x-\sqrt{3}|-\frac{1}{2 \sqrt{6}} \ln |\sqrt{2} x+\sqrt{3}|+C$$
Work Step by Step
Given $$\int \frac{d x}{2 x^{2}-3}$$
Since
\begin{aligned}
\frac{1}{2 x^{2}-3} &=\frac{1}{(\sqrt{2} x-\sqrt{3})(\sqrt{2} x+\sqrt{3})} \\
&=\frac{A}{(\sqrt{2} x-\sqrt{3})}+\frac{B}{(\sqrt{2} x+\sqrt{3})} \\
&=\frac{A(\sqrt{2} x+\sqrt{3})+B(\sqrt{2} x-\sqrt{3})}{2 x^{2}-3}\\
1 &=A(\sqrt{2} x+\sqrt{3})+B(\sqrt{2} x-\sqrt{3})
\end{aligned}
Then
\begin{align*}
\text{at } x&=\sqrt{3}/\sqrt{2} \ \ \ \ \to A=\frac{1}{2\sqrt{3}} \\
\text{at } x&=-\sqrt{3}/\sqrt{2}\ \ \ \ \to B=-\frac{1}{2\sqrt{3}}
\end{align*}
Hence
\begin{aligned}
\int \frac{d x}{2 x^{2}-3} &=\int \frac{\frac{1}{2 \sqrt{3}}}{(\sqrt{2} x-\sqrt{3})} d x-\int \frac{\frac{1}{2 \sqrt{3}}}{(\sqrt{2} x+\sqrt{3})} d x\\
&=\frac{1}{2 \sqrt{6}} \int \frac{\sqrt{2}}{(\sqrt{2} x-\sqrt{3})} d x-\frac{1}{2 \sqrt{6}} \int \frac{\sqrt{2}}{(\sqrt{2} x+\sqrt{3})} d x \\
&=\frac{1}{2 \sqrt{6}} \ln |\sqrt{2} x-\sqrt{3}|-\frac{1}{2 \sqrt{6}} \ln |\sqrt{2} x+\sqrt{3}|+C
\end{aligned}
