Question

Solve $\int \frac{1}{2x^2-3}dx=$

• A. $\frac{1}{\sqrt{6}}log\left|\frac{\sqrt{2}x-\sqrt{3}}{\sqrt{2}x+\sqrt{3}}\right|+c$
• B. $\frac{1}{2\sqrt{6}}log\left|\frac{\sqrt{2}x-\sqrt{3}}{\sqrt{2}x+\sqrt{3}}\right|+c$
• C. $\frac{1}{2\sqrt{6}}log\left|\frac{\sqrt{2}x+\sqrt{3}}{\sqrt{2}x-\sqrt{3}}\right|+c$
• D. $\frac{1}{\sqrt{6}}log\left|\frac{\sqrt{2}x+\sqrt{3}}{\sqrt{2}x-\sqrt{3}}\right|+c$

Answer 1

The correct option is B $\frac{1}{2\sqrt{6}}log\left|\frac{\sqrt{2}x-\sqrt{3}}{\sqrt{2}x+\sqrt{3}}\right|+c$

$\int \frac{1}{{2x}^2-3}dx$

$=\frac{1}{2}\int \frac{1}{x^2-\frac{3}{2}}dx$

$=\frac{1}{2}\int \frac{1}{{\left(x\right)}^2-\left(\sqrt{\frac{3}{2}}\right)}cdx$

$=\frac{1}{2}\frac{1}{2\times \sqrt{\frac{3}{2}}}\log \left|\frac{x-\frac{\sqrt{3}}{\sqrt{2}}}{x+\frac{\sqrt{3}}{\sqrt{2}}}\right|+c$

$=\frac{1}{2}\frac{1}{\sqrt{6}}\log \left|\frac{\sqrt{2}x-\sqrt{3}}{\sqrt{2}x+\sqrt{3}}\right|+c$

$=\frac{1}{2\sqrt{6}}\log \left|\frac{\sqrt{2}x-\sqrt{3}}{\sqrt{2}x+\sqrt{3}}\right|+c$

Hence, the answer is $\frac{1}{2\sqrt{6}}\log \left|\frac{\sqrt{2}x-\sqrt{3}}{\sqrt{2}x+\sqrt{3}}\right|+c.$