Integration Formula Question (1 Viewer)

[blackops23]

Hi guys, quick question:

I know the formula for integrating 1/(x^2 - a^2) = (1/2a)[ln(x-a) - ln(x+a)] + c

and for integrating 1/(a^2 - x^2) = (1/2a)[ln(a+x) - ln(a-x)] + c

---

So I was wondering, what is the formula for e.g. integrating 1/(a^2 - [f(x)]^2)??

Vice versa, formula for 1/([f(x)]^2 - a^2)??

does the numerator need to f'(x)??

E.g. integrate 1/(4-25x^2) (without substitution)

thanks appreciate the help

 

[Drongoski]

I can explain all this so easily - but not on bos because there are contraints of this medium.

 

[blackops23]

I can explain all this so easily - but not on bos because there are contraints of this medium.

But most people would not know what I'm talking about. So it's my loss not being able to have the joy of sharing my techniques with them just as it is their unwitting loss.

lol all you have to do is state what the formula is... Thanks anyways

 

[Trebla]

Honestly, you don't need to memorise any of these formulas. Firstly, because those types of integrals are rarely ever found in exams and secondly because even if they do pop up, they are easily derivable by partial fractions.

 

[Drongoski]

lol all you have to do is state what the formula is... Thanks anyways

No. That is not the way to do it. It is more an approach than a set of formulae.

 

[blackops23]

Honestly, you don't need to memorise any of these formulas. Firstly, because those types of integrals are rarely ever found in exams and secondly because even if they do pop up, they are easily derivable by partial fractions.

I know, I know, but partial fractions would take up an extra minute in an exam, and every second counts in a 4U exam...

 

[Trebla]

I'd much rather take that extra minute and derive it than commit it to memory for many reasons:
- Such questions are often worth several marks so by simply quoting the answer, you risk getting zero if it is incorrect for showing no working rather than partial marks. For some questions worth several marks, a 'bold answer' (is that spelt correctly?) generally does not warrant full marks (especially considering the formula is not within the syllabus). If you have read notes from the marking centre from past HSC exams, they tend to look favourably upon those who derive the results rather than quote a rote-learned result
- Such questions rarely appear in examinations, so it is really not worth all that time/effort commiting it to memory for an unlikely question (especially considering that you could answer it through partial fractions anyway in the event that it does appear) when you could have been covering other areas which are more likely to be examined

...and for the record, if you go ahead and let u = f(x) then the 'formula' should be obvious

 

[ferdin]

Hi guys, quick question:

I know the formula for integrating 1/(x^2 - a^2) = (1/2a)[ln(x-a) - ln(x+a)] + c

and for integrating 1/(a^2 - x^2) = (1/2a)[ln(a+x) - ln(a-x)] + c

---

So I was wondering, what is the formula for e.g. integrating 1/(a^2 - [f(x)]^2)??

Vice versa, formula for 1/([f(x)]^2 - a^2)??

does the numerator need to f'(x)??

E.g. integrate 1/(4-25x^2) (without substitution)

thanks appreciate the help

hi
use partial fraction:
1/4-25x^2 = A/(2+5x) + B/(2-5x)
A=1/4 and B=1/4
so the integral would be 1/20 Ln(2+5x)/(2-5x) + c

 

[cutemouse]

You could use hyperbolic functions. The standard integrals for them are easy to remember....

 

[seanieg89]

 

[blackops23]

...and for the record, if you go ahead and let u = f(x) then the 'formula' should be obvious

Thank you... the piece of gold I was searching for...

 

[Riproot]

I know, I know, but partial fractions would take up an extra minute in an exam, and every second counts in a 4U exam...

They usually get you to separate it into partial fractions in part i anyway...