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Factorising quadratic equations This method is used for the following factorable expression: ax 2 +bx+c Although this method is particularly useful with quadratic expressions with a≥2, it can be also used wh...

This is an integral many people struggle with, but, with a simple trick it becomes a little more straight forward. We will approach this integral using integration by parts. But what are the parts? Well, we ...

This will use the process of integration by parts. First, notice that ln(x)=ln(x)*1. So, the integral of ln(x) is the integral of ln(x) 1. The process of integration by parts is; int (v du / dx ) dx =vu - in...

This problem is all about using integration by parts, so let's start by quoting the formula for integration by parts: ∫u*(dv/dx)dx = uv - ∫v*(du/dx)dx To get the integral we want on the left hand side we can...

Usually the easiest way to find the root of a quadratic is to factorise it so you have something a form similar to - (x+2)(x-1)=0. From here you need to find a value for x that makes each bracket equal to ze...