Question

Integration of f(x) with respect to x where f(x) = $\frac{1}{{(x-\frac{1}{2})}^2+\frac{3}{4}}$ will be-

• A. $\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)$ + C
• B. $\frac{\sqrt{2}}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)$ + C
• C. $\frac{2}{3}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)$ + C

Answer 1

The correct option is A $\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)$ + C

The given problem is in the form of $\frac{1}{x^2+a^2}$ and we can directly use the corresponding formula which is $\int \frac{1}{x^2+a^2}$dx = $\frac{1}{a}ta{n}^{-1}\left(\frac{x}{a}\right)$ + C
Here, instead of simply x we have $(x-\frac{1}{2})$ and $a^2$ is equal to $\frac{3}{4}$. So the respective answer would be $\frac{2}{\sqrt{3}}ta{n}^{-1}\left(\frac{x-1/2}{\frac{\sqrt{3}}{2}}\right)$ + C
or $\frac{2}{\sqrt{3}}ta{n}^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)$ + C