Question

Integration of $f\left(x\right)$ with respect to $x$ where $f\left(x\right)=\frac{1}{(x-1/2{)}^2+3/4}$ 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. $\frac{2}{3}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+C$
• D. None of these

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 $\int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C$
Here, instead of simply $x$ we have $(x-1/2)$ and $a^2$ is equal to $3/4$. So the respective answer would be $\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{x-1/2}{\frac{\sqrt{3}}{2}}\right)+C$
or
$\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+C$