Partial fractions

[Stick]

During these holidays our teacher has given us a partial fractions handout to prepare us for integration early next term. We did this in GMA last year and I remember having all sorts of difficult problems such as having quadratic factors in the denominator that cannot be factorised or having to split up a fraction into three (and sometimes one answer would cancel out due to a zero on the numerator). However, all the questions I've seen in the Specialist course seem to be the really nice ones where we're either dealing with a factorisable quadratic factor or doing simple division of polynomials. Since partial fractions in itself is probably not going to be assessed, do I then assume that the partial fractions will intentionally remain simple, or have I just been exposed to simple material?

[Planck's constant]

I am almost 99.9% certain that you need to know the derivative of the inverse tan function, and recognise it for antidifferentation, therefore you cannot assume that you will not be tested on partial fractions with non-factorisable quadratics in the denominator.

[pi]

I am almost 99.9% certain that you need to know the derivative of the inverse tan function, and recognise it for antidifferentation, therefore you cannot assume that you will not be tested on partial fractions with non-factorisable quadratics in the denominator.

Agreed.

I wouldn't assume it's all going to be simple, but you might have to look beyond textbooks to find questions of a higher standard like this.

[Stick]

Awesome. I'll grab out my GMA textbook tomorrow and make sure to find the more difficult problems from last year.

[b^3]

Will just point out that they 'can' specifically test you on partial fractions (without linking it into anything else) in Specialist Maths, such as MC Q9, Exam 2 2007 http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2007specmaths2.pdf

[BigAl]

I remember a question in one of the trials where I used u du sub to get what was  like an ordinary fraction..and then I used the completing the square method in the denominator because it wasn't factorisable... And then taking the common factor out and using t dt sub to solve the problem. It was an indefinite integral so I first put back u and then x...plus c.