Which of the following is an anti-derivative of $$\frac{1}{4x^2 + 1}$$ ? A - HSC - SSCE Mathematics Extension 1 - Question 3 - 2020 - Paper 1

To find an anti-derivative of the function $\frac{1}{4x^2 + 1}$, we can recognize it as a standard integral formula related to the arctangent function:

$\int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$

In this case, we have:

• $a^2 = 4$ which gives $a = 2$

Thus, we can write the integral as:

$\int \frac{1}{4x^2 + 1} \, dx = \frac{1}{2} \tan^{-1}(2x) + C$

This indicates that the correct anti-derivative involves the expression $\frac{1}{2} \tan^{-1}(2x) + c$.