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There's a nice trick here you can do, treat the equation as 1*ln(x) then intergrate by parts.
Differentiating ln(x) gives 1/x, while intergrating 1 gives x
So your left with a much easier intergration
x*ln(x)-(Intergral sign)x* 1/x dx
which is simply x*ln(x)-xsee more
Well, as with all STEP questions, each problem is unique. But with graphing problems there are some nice things you can do to help break down the problem. Firstly, ALWAYS DIFFERENTIATE, i cannot stress this enough. If you know how your gradient is changing, this gives a lot about what a graph looks like. Secondly, look for roots to both you equation and its derivative. This is a BIG part of these questions and where the most marks will be awarded. Find where your graph hits the y and x axis, and find where the fixed points are. Finally, once you have all your fixed points, make sure to note if they are a maximum, minimum or a turning point.
Now, you should have a pretty good idea of how your graph looks, but there's still one more thing we can do. Find what your graph looks like as it tends to +/- infinity
For example as y = e^x - x^3 tends to infinity, e^x will increase much faster than x^3, so will look a lot like e^x for large x
But for large negative x, e^x is essentially 0, so the graph will look pretty much like x^3see more