## Calculus (3rd Edition)

$$\frac{1}{2 \sqrt{6}} \ln |\sqrt{2} x-\sqrt{3}|-\frac{1}{2 \sqrt{6}} \ln |\sqrt{2} x+\sqrt{3}|+C$$
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}