# Regarding the integration of rational functions

#### mentaaal

Joined Oct 17, 2005
451
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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#### Caveman

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

#### mik3

Joined Feb 4, 2008
4,843
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!
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?

#### Mark44

Joined Nov 26, 2007
628
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