The formula sheet Cambridge gives you during your exam tells you the following about this particular integral:
$$ \int \frac{1}{x^2+a^2}dx = \frac{1}{a}tan^{-1}(\frac{x}{a}) $$
This works for integrals of the form $\frac{1}{x^2+a}$ where $a$ is any rational number. You can still use this for integrals of the form $\frac{1}{bx^2+a}$, where $b$ is also a rational number; we just have to modify it a bit. What we’re looking for, in essence, is for the coefficient of $x^2$ to be $1$.
We’ll deal with $\frac{1}{4x^2+1}$ as an example.
$$ \int \frac{1}{4x^2+1}dx $$
Taking $4$ as common from the denominator:
$$ \int \frac{1}{4(x^2+\frac{1}{4})}dx $$
Now, we’ll take the constant “outside”:
$$ \frac{1}{4} \int \frac{1}{x^2+\frac{1}{4}} $$
We’ll integrate as normal from here:
$$ \frac{1}{4} \times [ \frac{1}{\sqrt{\frac{1}{4}}}tan^{-1}(\frac{x}{\sqrt{\frac{1}{4}}}) ] $$
$$ \frac{1}{4} \times [ 2tan^{-1}(2x)] $$
And we’ll end up with $\frac{1}{2}tan^{-1}(2x)$. Problem solved!