Question

Integrate ((x-1)/(x^5))^(1/2) dx show work please

Answer 1

\[\int\sqrt{\frac{x-1}{x^5}}dx\]
is this the question? this looks like an intense problem if it is correct...

Answer 2

yes

Answer 3

have you learned trigonometric transformation?

Answer 4

No, though I don't think this problem is that complicated....I have an answer written out in the answers section of the book (Thomas Calculus) but Im still a little confused as to the mechanic of solving.

Answer 5

I can only think of one way of doing this, and I dont know that it would actually work.

Answer 6

This problem solved using substitution.

Answer 7

Hold on a second, will post a screenshot of what the book says.

Answer 8

yeah... i know i sub'd then trig sub'd in my head--and then it go to where im not sure if it would work.

Answer 9

check that out

Answer 10

that solution makes no sense at all.

Answer 11

well... i guess it makes sense for x>0

Answer 12

\[\sqrt{\frac{x-1}{x^5}}=\frac1{x^2}\sqrt{\frac{x-1}{x}}\]
for x>0

Answer 13

What confuses me is where they would get that U from...

Answer 14

would it make sense to set the entire equation inside the square root to U?

Answer 15

well, when you are doing those substitutions you pick your own u and find its du.
They defined their u as being whats under the radical towards the end. I have a problem with the algebra, after step 2 it's pretty straightforward and correct.

Answer 16

what don't you like about the algebra?

Answer 17

what I said above, it only holds for positive values of x.

Answer 18

really?

Answer 19

really.

Answer 20

I get then being equal for negative values

Answer 21

them

Answer 22

there is nothing wrong with the above solution

Answer 23

Zarkon, could you write out your work please?

Answer 24

you have the work

Answer 25

Where does the u = 1-1/x come from?

Answer 26

that is what they picked for u...since du will get rid of the \(1/x^2\)

Answer 27

a more basic version would be to say:
\[\sqrt{x^5}=x^{5/2}\]
Let's take x=-6
\[\sqrt{-6^5}=-6^{5/2}\]
\[36i\sqrt{6}=-36\sqrt{6}\]
one is real, the other isnt--this is a false simplification.

Answer 28

\[\sqrt{x^5}=\sqrt{x^4x}=x^2\sqrt{x}\]

Answer 29

again...what they did was correct

Answer 30

try graphing both functions

Answer 31

Wolfram seems to think this is true only for x>0 as well.
I think perhaps this is coming back to parenthetical insertions.

That is--if you replace x with a number in parenthesis it works.

http://www.wolframalpha.com/input/?i=%28%28x-1%29%2F%28x%5E5%29%29%5E%281%2F2%29+%3D+1%2Fx%5E2+%28%28x-1%29%2Fx%29%5E%281%2F2%29+

Answer 32

the point is \( \sqrt{x^4} = x^2 \) for all x.

Answer 33

@JamesJ

only with parenthetical insertions (which I was taught not to do...)

Answer 34

let x=-1

\[\sqrt{\frac{x-1}{x^5}}\]
\[\sqrt{\frac{-1-1}{(-1)^5}}=\sqrt{-2/(-1)}=\sqrt{2}\]

\[\frac1{x^2}\sqrt{\frac{x-1}{x}}\]

\[\frac1{(-1)^2}\sqrt{\frac{-1-1}{-1}}=1\cdot \sqrt{-2/(-1)}=\sqrt{2}\]

Answer 35

I have no idea what parenthetical insertions are, but I imagine that if a bunch of math geeks got drunk, it would be very funny.

Answer 36

I wish I could talk more ... but I'm going out to eat...I'll be back though ;)

Answer 37

it's what Zarkon just showed.

for instance with
(x^4)^(1/2)=x^2

If we insert our negatives in parenthesis it works: x=(-4)
((-4)^4)^(1/2) = (-4)^2
256^(1/2)=16
16=16

However, if we don't ..because you are adding a new, higher order, step to evaluate when you do: x=-4

(-4^4)^(1/2) = -4^2
(-256)^(1/2) = -16
16 i = -16

Answer 38

From
"(-4^4)^(1/2) = -4^2"
to
"(-256)^(1/2) = -16"

is false. because, for instance, you're evaluating (-4)^2, which is 16, not -16;
or (-4)^4 = 256, not -256

Answer 39

i dunno maybe after complex analysis my brain just became fried when it came to exponents and such

Answer 40

^^ James thats getting back to exactly what I was talking about.

(-4)^2 =/= -4^2

Order of operations ;)

Answer 41

It's not even a question of order of operations. In any case, it is true for all real numbers x that
\[ \sqrt{x^4} = x^2 \]
and the reworking of the integrand is correct.

Answer 42

So, are you saying that

(-4)^2 = -4^2
?
Really?

Answer 43

No. Consider the following expression
\[ \sqrt{(-x)^4} \] We can write -x as =(-1)x. Hence
\[ \sqrt{(-x)^4} = \sqrt{ (-1)^4x^4} = \sqrt{ 1.x^4} = \sqrt {x^4} \]

Therefore \( \sqrt{(-x)^4} = \sqrt{x^4} \) and that expression is equal to \( x^2 \) for all real numbers \( x \).

Answer 44

Anyway Frank, do you have your answer or not?

Answer 45

what a crazy thread

Answer 46

really.