Question

Integration!
\[\int \frac{x-3}{x^2 + 2x + 2} dx\]

Answer 1

I couldn't see a u that works as-is.

I'm tried completing the square to see if that did anything productive.

\[\int \frac{x-3}{(x+1)^2 +1}dx\]\[u = x + 1 \\ du = dx\]
So.
\[\int \frac{u - 4}{u^2 + 1} du\]
That's a little better.

Answer 2

Hmm have you learned trig sub's yet?
It works quite nicely for this problem :o

Answer 3

I know them in theory. In practice I'm a bit shaky. Hold my hand and I'll give it a shot? :D

Answer 4

Actually your u sub works our really well also,\[\Large \int\limits\limits \frac{u - 4}{u^2 + 1} du\quad=\quad \int\limits\limits \frac{u}{u^2+1}du-4\int\limits \color{orangered}{\frac{1}{u^2+1}}du\]

Answer 5

Do you remember that `orange` integral by chance? :)
It's a good one to memorize!

Answer 6

No, but I have a table of integrals close at hand and that's one of those pretty ones that is surely on it. :D

Answer 7

So we split it up into a couple of parts.
For the first part, you can apply another substitution if it's not jumping out at you yet :)
m=u^2+1 should get the job done.

Answer 8

\[v = u^2 + 1 \\ dv = 2u du\]\[\frac{1}{2} \int \frac{1}{v} dv = \ln |v| = \ln |u^2 + 1| = \ln |(x+1)^2 + 1| + C\]
Yes, that's not so bad.

So combining that with ye olde integral table I get:
\[\frac{\ln |x^2 +2x + 2|}{2} - 4\tan^{-1}(x + 1) + C\]

Answer 9

Yay good job \c:/

Answer 10

Out of curiosity, what would be the trig sub?

Answer 11

I guess I can kind of work backwards from my answer. Something involving tan!

Answer 12

After completing the square as you did,\[\Large \int\limits \frac{x-3}{(x+1)^2 +1}dx\]We could utilize a trig sub to get rid of the nasty addition in the denominator.
We would let,\[\Large x+1\quad=\quad \tan \theta\]\[\Large x-3\quad=\quad \tan \theta-4\]\[\Large dx\quad=\quad \sec^2\theta\;d \theta\]Those are the pieces we would use for our substitution.

Answer 13

\[\Large \int\limits\limits \frac{x-3}{(x+1)^2 +1}dx\qquad\to\qquad\int\limits\limits \frac{\tan \theta-4}{(\tan \theta)^2 +1}(\sec^2\theta \;d \theta)\]

Answer 14

See how our trig identity will get rid of that nasty addition in the bottom? :)

Answer 15

Yes!

Answer 16

\[\int \frac{\tan \theta - 4}{(\tan \theta)^2 + 1} \sec^2\theta d\theta\]\[\int \frac{\tan \theta - 4}{\sec^2 \theta} \sec^2\theta d\theta\]\[\int (\tan \theta -4) d\theta\]

Answer 17

Yah! It simplifies down quite nicely huh?
Tangent is another one of those weird integrals that you might want to just memorize.
You can change it to \(\Large \dfrac{\sin\theta}{\cos\theta}\) and do a u-sub from there if you want though.

Answer 18

Ok. So.\[v = \cos \theta \\ dv = -\sin \theta\]
\[- \int \frac{1}{v} dv = -\ln |v| + C = -\ln |\cos \theta| + C\]
Plus the \(4 \int d\theta\) leaves \(4\theta - \ln |\cos \theta| + C\)

Ok, so now how to get x back.

\[x + 1 = \tan \theta\]\[x + 1 = \frac{\sin \theta}{\cos \theta}\]
Er.. I can solve that for \(\cos \theta\) but there is still a theta floating around.

Answer 19

Oh, the 4 was minus. Stupid details. :D
But anyway.

Answer 20

I can see from comparing my two answers that somehow
\[\frac{\ln |x^2 + 2x + 2|}{2} + C= -\ln |\cos(\tan^{-1}(x+1)))| + C\]
But hopefully I can be forgiven for not grasping the immediate relationship.

Answer 21

Oh yes there is another annoying step when doing trig subs.
Undoing the substitution is a bit tricky.

We have to go back to good ole triangle math! :)\[\Large \tan \theta\quad=\quad x+1\qquad=\quad\frac{x+1}{1}\quad=\quad\frac{opposite}{adjacent}\]

Answer 22

We draw a triangle, label one of the angles theta, and then label the sides according to our relationship.

Answer 23

Oh!
How geometric.

So \(\cos \theta = \frac{1}{\sqrt{(x+1)^2 + 1}}\)
Among other things.

I think Pythagoras is making it worse?

Answer 24

Worse or better .. I dunno.
But it's leading us to the same answer we got using the other method! :O

Answer 25

Oh!
\[-\ln |(x^2 + 2x + 2)^{-1/2}|\]

Answer 26

Yah there ya go! :D

Answer 27

Now I see it! Thank you. :D

Answer 28

yay team c:

Answer 29

It's quite clear now. Thank you for the detailed explanations.
I would give you another medal if I could. :P :D