**Q: ∫1 / (u²+ 2) du**

A: When looking at this problem, I think the first step is trig. substitution. Why do I think that? Experience.

I remember the two most important trig identities when considering trig substitution in intergration:

(1) sin²x + cos²x = 1

(2) tan²x + 1 = sec²x

Since I see u² + 2 in my original problem, I think that this looks more like (2)… Therefore, I try the following:

Let u = √(2)*tanx

Why did I also include √(2)?? Watch and see:

If u = √(2)*tanx then du = √(2)*sec²x dx

Now, substitute in u and du into our orignal equation, like so:

∫1 / (u²+ 2) du

∫1 / ((√(2)*tanx)²+ 2) [√(2)*sec²x dx]

Do the colors help to show where I have made the substitutions, or do they just make it more confusing?

Anyhow, now for some house keeping (square things out, multiply my fractions, etc…):

∫1 / ((√(2)*tanx)²+ 2) [√(2)*sec²x dx] = ∫1 / (2*tan²x+ 2) [√(2)*sec²x dx]

= ∫√(2)*sec²x / (2*tan²x+ 2) dx

Now, I have one fraction. In the bottom, notice both constants are “2s”? Thanks to my intial choice of √(2)! So, factor out a 2:

∫√(2)*sec²x / [2(tan²x+ 1)] dx

Wait… Remember that trig identity from above? tan²x + 1 = sec²x

Therefore, replace tan²x + 1:

∫√(2)*sec²x / (2sec²x) dx

Aha. Now the secants cancel!

∫√(2)/2 dx

We are just left to integrate a constant:

∫√(2)/2 dx = √(2)/2 * (x) + C

Well, we are almost done. Since the orignal problems were given to us in terms of “u”, we better put it back into u’s and not x’s.

Recall: u = √(2)*tanx. Solve for x:

u = √(2)*tanx

u / √(2) = tanx

arctan [u / √(2)] = x

Therefore, our answer: √(2)/2 * (x) + C = √(2)/2 * (arctan [u / √(2)]) + C

We have not even covered trig sub yet, I’ll post the work I did and lets see if we can work through it, maybe the answer I have is the same I don’t know, I’ll do that in a minute.

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This is the work that I did I got close to the right answer but of course close is only good in horseshoes and hand grenades lol.Integrate {(1)/(x^2+2x+3)} dx

Check discriminant: b – 4ac <= 0 than complete square

Discriminant = -10 so complete the square

(x+1)^2 + 2 is the new denominator

Let u = x + 1

so du = dx

We then get the following:

Integrate {1/[(x + 1)^2 +2]} dx

= Integrate {1/(u^2 + 2)} du

Int. {1/2(u^(2)/2 + 1)}= 1/2 Int. {1/{(u/ radical 2)^2 +1]}

————————————————

= sqrt(2)/2 Int. {1/(u^2 + 1)}

= sqrt(2)/2 arctan(u) + constant

= sqrt(2)/2 arctan(x+1) + constant

Above is the work I did for the same problem that you did above, are they same and is what I did remotely correct?

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Everything in BOLD up until the “—————————” I added is good (I edited your comment)

The very next line, where a sqrt(2) is factored out has gone wrong.

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I took like this:

(U/sqrt(2))^2 and moved the radical 2 from under the U to the numerator of the 1/2 outside the radical sign is that incorrect? I tried to follow some notes that I had from class

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I will pick up where things went south:

= 1/2 Int. {1/{(u/ radical 2)^2 +1] du }

Immediately from here, we can see that “arctan” will be in our answer….

Now, another substitution. Let w = u/sqrt(2), dw = 1/sqrt(2) du

So, du = sqrt(2) dw

= 1/2 int { 1 / (w^2 + 1) sqrt(2) dw}

Clean house:

= sqrt(2)/2 int { 1 / (w^2 + 1) dw}

= sqrt(2)/2 * arctan(w) + C

= sqrt(2)/2 * arctan(u/sqrt(2)) + C

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Aaahhhhhh I see another substitution was needed to do it correctly all I was missing was that stinkin radical 2 in the denomicator of the arctan always the small stuff lol. thanks so much Stacy, this is gettin tweeted.

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Response to comment #4:

Yes, this is incorrect because you are factoring out something from a denominator and placing it in a numerator — it just changes spots too magically. Also, since the [(u/sqrt2)^2 + 1] is bound together by a plus sign, you must be factoring numbers out of BOTH of them in order to pull anything out…. Make sense?

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yeah totally makes sense I will have to go through my notes again because I don’t remember doing anything like that, maybe the prof. made a mistake? lol that would be extra points on the exam haha

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My first thought was just to use the arctan integration formula, with a=sqrt(2). It works a little (a lot) quicker – but now that I’m also teaching calculus 2 it’s nice to see how trig substitution will also work!

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Hi Michelle,

Yes, using the arctan integration formula is great if you either have it memorized or are able to use a book for reference. But, in the event that this shows up on a test and you have to figure it out from scratch it is very nice to know the trig substitution method or the other substitution methods!

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