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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...

For any right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is the sum of the square of the remaining two sides.

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...

The integral of log(x) is not necessarily straight-forward. Though we can use the fact that d/dx(log(x)) = 1/x to help us. Rather than simply trying to integrate log(x), we can use integration by parts on 1 ...

The key to solving any integral of this form is to use the cosine rule: cos(2x) = cos 2 (x) - sin 2 (x) = 2 cos 2 (x) - 1 = 1 - 2 sin 2 (x) All of these forms are really helpful when solving problems such as...