# Regarding the integration of rational functions

Discussion in 'Math' started by mentaaal, May 5, 2008.

1. ### mentaaal Thread Starter Senior Member

Oct 17, 2005
451
0
Hey guys have a wee problem with rational problems.

Its actually a letter to my lecturer but well... he probably wont see it for a few days.

Cheers!

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2. ### Caveman Senior Member

Apr 15, 2008
471
1
What are you proposing the answer is to the arctan method?

3. ### mik3 Senior Member

Feb 4, 2008
4,846
67
i dont see the reason to square them. The answer is simply 14ln(4+t)+C.
And you dont really need to use substitution to find this. because the numerator is the derivative of the denominator if you take 14 out of the integral, because is a constant, you can say directly that the result is 14ln(4+t)+C
why do you want to complicate things?

4. ### Mark44 Well-Known Member

Nov 26, 2007
626
1
Greg,
For the sake of simplicity in using LATeX, let me simplify the second integral you have, so that the rational function is 1/(t$^{2}$ + 1).

The problem is that the integral you have with $\sqrt{t}^{2}$ + 1$^{2}$ in the denominator isn't equal to tan$^{-1}$($\sqrt{t}$). (Yes, I know your integral is slightly different.)

To verify this statement, take the derivative of the inverse tangent expression. If you do, you'll find that the derivative has a (t + 1) factor in the denominator, but there is also a factor of 2$\sqrt{t}$. Since you didn't get back to your integrand when you differentiated the antiderivative, then what you thought was the antiderivative actually wasn't.

The reason is this:
When you replaced t + 1 with $\sqrt{t}^{2}$ + 1, you have changed the variable of integration, but you didn't change the differential part: you still have dt. If you had instead, d($\sqrt{t}$), it would have worked.

The bottom line here is that when you use integration formulas, you have to make sure that the integrand your are working on matches exactly the one in your formula, including the differential part.

BTW, in your PDF, you're addressing the instructor as both Brendan and Brenda...
Mark

5. ### mentaaal Thread Starter Senior Member

Oct 17, 2005
451
0
Mik, the reason why i am asking is not to complicate matters, it is by asking questions like these that i obtain a better understanding of what is going on. If i merely accept which rules to use in certain circumstances and not not question the reason then I would find it harder to apply them.

And thanks mark for the excellent response. I find calculus is branch of maths in which it is very easy to make a blunder!

Ha ha I am sure my lecturer will appreciate my calling his gender into question.

Last edited: May 5, 2008