Question

integrate :

dx/square root 28-12x+x^2

Answer 1

\[\int{dx\over28-12x+x^2}\]?

Answer 2

i think there should be a swrt

Answer 3

\[dx \div \sqrt{28-12x+x^2}\]

Answer 4

use partial fraction method.....should work

Answer 5

oops\[\int{dx\over\sqrt{28-12x+x^2}}\]

Answer 6

hmm looks like trig sub though

Answer 7

definitely trig sub

Answer 8

yes thats right Turing test
its integration by parts

Answer 9

arcsin?

Answer 10

oh by parts

Answer 11

What if we write the bottom as (x-6)^2 - 8 and then u = x - 6?

Answer 12

wow again? lol

Answer 13

i've answered 2 integration by parts today already >_>

Answer 14

So we are left with \[\int \frac{dx}{\sqrt{(x-6)^2 - 8}} = \int \frac{du}{\sqrt{u^2 - 8}}\]

Answer 15

..and how is that not a trig sub?

Answer 16

Afterwards, yes, I agree. So disregard.

Answer 17

I was trying to imagine a PF way to do it... got me all interested haha

Answer 18

@roselynECE do you know what substitution to make?

Answer 19

this is not substitution its integration by parts

Answer 20

are you sure about that?

Answer 21

yes

Answer 22

integrals with fractional exponents don't usually involve integration by parts unless it's got a logx with it or something, so good luck to whoever figures out how to do this by parts!

Answer 23

i see inverse trig

Answer 24

me too

Answer 25

...which can be shown with a trig sub...

Answer 26

Im sorry for that I was just overlook

Answer 27

no apologies necessary :)

Answer 28

Haha @TuringTest sorry for that. At first glance, I thought I could expand it out into PF. But, yeah, hopes were gone.

Answer 29

how to solve this using arcsin?

Answer 30

What I did from above is to say that u = sqrt(8)*sec(v) and du = sqrt(8)*tan(v)sec(v)dv but I suppose there are easier ways.

Answer 31

if you use completing the square...youll get

\(\LARGE \int \frac{dx}{\sqrt{(6-x^2)^2 - 8}}\)

Answer 32

that's (6-x)^2 in the denom

Answer 33

draw a triangle...let the hypotenuse = 6- x.base leg = \(\sqrt 8\) and the other leg = \(\sqrt{(6-x)^2 - 8}\)

so let \(\sqrt 8 \sec \theta = 6-x\)
\(\sqrt 8 \tan \theta = \sqrt{(6-x)^2 - 8}\)
\(\sqrt 8 \sec \theta \tan \theta d\theta = -dx\)

substitute these values to your integral

Answer 34

can you tell me what you have from substituting those into your integral @rocelynECE

Answer 35

@roselynECE sorry misspelled last time