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### Course: AP®︎/College Calculus AB>Unit 6

Lesson 12: Integrating functions using long division and completing the square

# Integration using completing the square and the derivative of arctan(x)

Sometimes we can integrate rational functions by using the method of completing the square in the denominator and then integrating using u-substitution and our knowledge about the derivative of arctan(x).

## Want to join the conversation?

• this was amazingly complicated compared to all the preceding videos
• I kind of agree; it seems complicated from the perspective of Sal not really discussing integrals of the inverse trig functions. When he said "... recognize arctan..." I was pretty sure I had not seen arctan in this context. If someone thinks I've missed something, please let me know.
• Why do they never give examples or questions using the anti-derivative or arccos(x) in the next skill?

I want to be prepared of that in case someone were to ask me, or if I were to ever need the information myself.

- ncochran2
• By "anti-derivative", I'm assuming you mean the antiderivative of the inverse trig functions? Well, here they are:

`∫ arcsin x dx = x arcsin x + √(1 - x²) + C`
`∫ arccos x dx = x arccos x - √(1 - x²) + C`
`∫ arctan x dx = x arctan x - ½ln(1 + x²) + C`
`∫ arccot x dx = x arccot x + ½ln(1 + x²) + C`
`∫ arcsec x dx = x arcsec x - ln(x + √(x² - 1)) + C`
`∫ arccsc x dx = x arccsc x + ln(x + √(x² - 1)) + C`

Now I don't know whether you're looking at the course from the general calculus page or the AP Calc AB / BC page, but I know that this content will not be on the AP Exam (if you're in the AP course).

As for arccos, arccsc, and arccot, I wondered about that myself previously, and I found that it was simply due to mathematicians choosing the convention that arcsin, arcsec, and arctan were to be used for the integrals, as the other three are simply the negative of arcsin, arcsec, and arctan respectively.
Hope that I helped.
• Is there a general pattern to look out for to know when completing the square will lead to an expression similar to trig derivatives?
• Only the arc trig functions' derivatives are numerical. To spot these within integrals, I look for the pattern a^2 + b^2 or a^2 - b^2. If there is a + sign between the terms, the integral is likely to evaluate to something with either arctan or arccot. If there is a - sign instead, the result of the integral is likely to involve arcsin or arccos.

Completing the square always results in one of those patterns.. so if there is something that resembles a quadratic in the denominator of your integral, then you end up with an integral that requires a process similar to that of the video's.
• Ive gone through all the Calc videos before this. Never once have we covered derivative of arctan(x).
• You don't need to know the derivative of `arctan(x)`. Just do a simple substitution `u = tan t`
``∫1/(u^2 + 1) du= ∫sec^2(t) dt/(1 + tan^2(t))= ∫dt= t + C= arctan(u) + C``
(1 vote)
• You don't need all that extra stuff to get your u. Much simpler to simply say: at 1/(x-3)^2 + 2^2, make u = x - 3. Then, when you take your antiderivative, you would get 1/5 (1/2 arctan (u/2) + C); distribute and unsubstitute for: 1/10 arctan (x+3)/2 + C.
• wha just hapend at ? why do we get ∫1/((x-3)/2^2)+1dx
• Sal divided everything by 4, so naturally, the 2^2 will be 1.
• at , why do we multiply the nominator by half?
• du=.5dx, so in order to substitute the existing 1dx with du we must first multiply it by .5 -> multiply the nominator by half.
• Is using ln(x) another option to integrate this rational function instead of using arctan(x)? And how would you differentiate between the two cases? Or are they just two ways to do the same problem?

For instance, with the problem Sal showed in the video, if you took the antiderivative as ln(x), you'd end up with ln(x^2-6x+13)/10(x-3), unless I'm mistaken. Is this equivalent to the result Sal got?