Accidently uploaded 2 of the same pictures, it wouldnt let me delete it 
@Lara.C, basically the goal is recognise the numerator is the differential of the denominator, which when you integrate gives a log.
Integration by substitution is an MX1 concept and not examinable here. They're required to use the rule \( \int \frac{f^\prime(x)}{f(x)}\,dx = \ln [f(x)]+C \).
Just wondering if there is a general rule for integrating fractions? It doesn't seem to be in my textbook and I seem to hit a deadend every time I run into a similar sort of problem.
E.g integral of (x/ (x^2 - 9))
Or integral of (1/(x+4))
I know how to for exponentials or logarithmic equations, but can't seem to make it work for these ones!
Thankss!! 
For the second one, integrals of the form \( \int \frac{1}{ax+b}dx \) you're just expected to know that turns into \( \frac{1}{a}\ln(ax+b)+C \)
Whereas for the first one, you need to get used to using that rule. Which relies essentially on you recognising that what's on the top, looks very similar to what's on the bottom, except it's out by a constant. In particular, the derivative of the bottom here is \( \frac{d}{dx}(x^2-9)=2x\), but on the numerator we just have \(x\) by itself.
...So we multiply some fudge factors to force the numerator into being \(2x\).
\begin{align*}\int \frac{x}{x^2-9}\,dx &= \frac12\int \frac{2x}{x^2-9}\,dx\\ &= \frac12\ln (x^2-9)+C\end{align*}
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