Question

@xapproachesinfinity

Answer 1

\[\int\frac{(x^2 + 2x+4)}{\sqrt{x^2-4x}}dx\]

Answer 2

alright, so we are going to do that trig substitution

Answer 3

so we need to draw our triangle

Answer 4

ok first u sub
u=x-2 du=dx
\[\int \frac{(u+3)^2+3}{\sqrt{u^2-4}}du\]
so \[u=2\sec\theta \Longrightarrow du=2sec\theta \tan \theta d\theta\]

Answer 5

yeah we the triangle but how did you get that part
?

Answer 6

what kind of trig sub did you do

Answer 7

I'm trying to remember calc I haven't had in like 5 years, feel free to correct me if wrong

Answer 8

ok let's go with the trig from what i did:)

Answer 9

agreed :)

Answer 10

so what do we get

Answer 11

so we get
\[\int \frac{(2sec\theta+3)^2+3}{2sec\theta}2secθtanθd\theta\]

Answer 12

?

Answer 13

yes!
we cancel you should get 2tan on the bottom not sec

Answer 14

other stuff are good

Answer 15

sec^2-1=tan^2 (we are using this identity under the radical :)

Answer 16

wait, I don't see the 2 tan on bottom

Answer 17

I only see it on top

Answer 18

\int \frac(2sec\theta+3)^2+3)tanθd\theta

Answer 19

well it was (2sec)^2-4=4sec^2-4=4(sec^2-1)=4tan^2
with root we get 2tan

Answer 20

but then how are we canceling?

Answer 21

cancel tan with tan
sec we need it

Answer 22

I can now see the tan goes away right?

Answer 23

\[\int [(2\sec\theta +3)^2]\sec\theta d\theta \] we get this

Answer 24

yea

Answer 25

yes exactly

Answer 26

I follow, but you dropped the second +3

Answer 27

oh yeah forgot that part :)

Answer 28

ok, so foild then we get a sec^2 which is tan, a sec, and the secant ^3 is doable

Answer 29

I see now you evil mastermind!

Answer 30

yeah we get sec^3 , sec^2 all of them are doable
sec^3 is by part

Answer 31

yep yep, alrighty, I got it. Thanks! I totally forgot about this technique existing!

Answer 32

I didn't just want to give that person the answer though

Answer 33

no problem!

Answer 34

to satisfy my own curiosity

Answer 35

:)

Answer 36

Thank you!! You are the sensei. I bow to your prowess

Answer 37

no my friend
you are way better, i saw some of your responses before

Answer 38

the sensei is you haha

Answer 39

:) we all have our moments. Some are more rare than others. You are always on point in calc. You get the sensei of calculus award :)

Answer 40

that should be a title haha

Answer 41

haha okay
thought i find alot of hurdles in calc
but it is supposed to be that way otherwise no point in learning it or any other math stuff

Answer 42

though*

Answer 43

true, wait till you have to prove calc. It's a nightmare

Answer 44

yeah i suppose but that the beauty of it!
alright go to go

Answer 45

alright, thanks again! Night!

Answer 46

got to go* my typing is horrible haha
alright! good night :)