The formula sheet Cambridge gives you during your exam tells you the following about this integral: \(\int \frac{1}{x^2+a^2} dx = \frac{1}{a}tan^{-1}(\frac{x}{a})\).
This form works for integrals of the form \(\frac{1}{x^2+a}\). You can still use this for integrals of the form \(\frac{1}{ax^2+b}\). You just have to modify it to fit those constraints. What we're looking for, in essense, 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:
\(\frac{1}{4} \int \frac{1}{x^2+\frac{1}{4}}dx\)
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 \(\frac{1}{2}tan^{-1}(2x)\).
This article was written on 2024-09-21. If you have any thoughts, feel free to send me an email with them. Have a nice day!