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The first thing to notice here is that the denominator of the integrand is a case of 'difference of two squares'. The integral, which I will call I, can be rewritten as the integral of 4/((1+x)(1-x)) dx. ...

Firstly we remind ourselves of the integration by parts formula. The formula used to integrate u dv/dx with respect to x = uv - the integral of v du/dx with respect to x.
So the first thing we have t...

The key fact here is to remember the relationship between the gradients of lines which are parallel and perpendicular. Two lines that are parallel will always have the same gradient. So if we have a line ...

Initially this looks unlike all the other differentiation questions and seems unsolvable. However the expression 2^x can be rewritten in an equivalent form that will allow us to use the differentiation ru...

Integration by parts can be considered as the inverse method of differentiation using the product rule. With the product rule we have: d(fg)/dx = f(dg/dx) + g*(df/dx) where f and g are functions ...