- #1

- 31

- 0

I can get the problem down to [tex]\int_1^8 9x^{2}+\frac{1}{\sqrt{2}}+\frac{1}{144x^{2}}} dx[/tex] , but I don't know how to integrate the last two terms in that integral. Am I doing something wrong?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter 7yler
- Start date

- #1

- 31

- 0

I can get the problem down to [tex]\int_1^8 9x^{2}+\frac{1}{\sqrt{2}}+\frac{1}{144x^{2}}} dx[/tex] , but I don't know how to integrate the last two terms in that integral. Am I doing something wrong?

- #2

VietDao29

Homework Helper

- 1,423

- 3

I can get the problem down to [tex]\int_1^8 9x^{2}+\frac{1}{\sqrt{2}}+\frac{1}{144x^{2}}} dx[/tex] , but I don't know how to integrate the last two terms in that integral. Am I doing something wrong?

How did you get it down to [tex]\int_1^8 9x^{2}+\frac{1}{\sqrt{2}}+\frac{1}{144x^{2}}} dx[/tex]? It doesn't look correct to me. Can you show us the steps so that we can check them for you?

Hint: Since there's a big square root, one should first try to Complete the Square.

- #3

- 31

- 0

I distributed the (1/2) power among everything under the radical. Is that incorrect?

- #4

SammyS

Staff Emeritus

Science Advisor

Homework Helper

Gold Member

- 11,368

- 1,035

Is [tex]\sqrt{1+1+1}=\sqrt{1}+\sqrt{1}+\sqrt{1}\ ?[/tex]I distributed the (1/2) power among everything under the radical. Is that incorrect?

I so , then [tex]\sqrt{3}=3\,.[/tex]

- #5

eumyang

Homework Helper

- 1,347

- 10

I distributed the (1/2) power among everything under the radical. Is that incorrect?

Yes, that's completely wrong. You can't do that with a sum inside a square root.

I would instead find a common denominator and add the three fractions.Hint: Since there's a big square root, one should first try to Complete the Square.

- #6

VietDao29

Homework Helper

- 1,423

- 3

I distributed the (1/2) power among everything under the radical. Is that incorrect?

You seem not to know how to Complete the Square.

Roughly speaking, Complete the Square is the manipulation you do to change your expression into something like: (a

Say, your expression is:

[tex]4x ^ 2 + 8x + 8[/tex]

Here's how you would do it. You'll try to split your expression into a square. Note that:

(*a* + *b*)^{2} = *a*^{2} + 2*ab* + *b*^{2}

[tex]4x ^ 2 + 8x + 8 = (2x) ^ 2 + 8x + 8 = (2x) ^ 2 + 2.(2x).2 + 8 = \left[(2x) ^ 2 + 2.(2x).2 + 2 ^ 2 \right] + 4 = (2x + 2) ^ 2 + 4[/tex].

Another example is:

[tex]x ^ 2 + 5 + \frac{1}{4x ^ 2} = x ^ 2 + \left( \frac{1}{2x} \right) ^ 2 + 5 = x ^ 2 + 2.x.\frac{1}{2x} + \left( \frac{1}{2x} \right) ^ 2 + 4 = \left( x + \frac{1}{2x} \right) ^ 2 + 4[/tex].

Let's see if you can Complete the Square for:

[tex]9x^{4}+\frac{1}{2}+\frac{1}{144x^{4}}[/tex] :)

- #7

- 31

- 0

I do not know how to complete the square since x is in the denominator. I can get it to [tex](3x^{2})^{2}+(\frac{1}{12x^{2}})^{2}= -\frac{1}{2}[/tex]

- #8

VietDao29

Homework Helper

- 1,423

- 3

The identity we should use here is:

(*a* + *b*)^{2} = a^{2} + **2***ab* + *b*^{2}

In your problem:

3

1/(12

So, what should you add to make it becomes (

You can look again at my 2 examples to have some ideas.

- #9

- 31

- 0

So 2ab is 1/2.

Then if we have the square root of (a+b)^2, that equals (a+b), and b still has x in the denominator, so how do I integrate that?

- #10

eumyang

Homework Helper

- 1,347

- 10

Rewrite

[tex]\frac{1}{12x^2}[/tex]

as an expression with a negative exponent, and then use the power rule.

- #11

- 31

- 0

Wow. I'm really off of my game tonight. I finally got it worked out. Thank you everyone for your help. I truly appreciate it.

Share: