Evaluate an Integral Using a Specific Hint

In summary, the conversation is discussing how to solve the integral of dx/(x4 + 16). The conversation covers various approaches, including using the given hint and splitting into partial fractions. The final solution is to create a system of equations by equating coefficients and solving for the constants. The conversation also includes a discussion on the difficulties and potential errors in solving the system.
  • #1
LilTaru
81
0

Homework Statement



[tex]\int[/tex] dx/(x4 + 16)


Homework Equations



Hint: With a>0, x4 + a2 = (x2 + [tex]\sqrt{}2a[/tex]x + a)(x2 - [tex]\sqrt{}2a[/tex]x + a)

The Attempt at a Solution



I've plugged this into the equation, which leaves me with:

[tex]\int[/tex] dx/[(x2 + [tex]\sqrt{}2a[/tex]x + a)(x2 - [tex]\sqrt{}2a[/tex]x + a)]

But now I am sooooo confused! Where do I go from here? This hint does not seem so helpful... Please help!
 
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  • #2
You could try splitting it into partial fractions now, which may make it easier.
 
  • #3
How would I do that?
 
  • #4
Rewrite the 1/hint

as


[tex] \frac{1}{{(x^2+ \sqrt{2a}x + a}{( x^2 - \sqrt(2a) + a)}} = \frac{Ax+B}{(x^2+ \sqrt{2a}x + a} + \frac{Cx+D}{ x^2 - \sqrt(2a) + a}[/tex]


then bring the right side to the same denominator and equate the numerators. Then choose values of x to get the constants A,B,C,D.
 
  • #5
Yeah. This was my first approach but if you plug in the value of a = 4, it seems the two factors of the hints never equal 0... Maybe this is just me.
 
  • #6
LilTaru said:
Yeah. This was my first approach but if you plug in the value of a = 4, it seems the two factors of the hints never equal 0... Maybe this is just me.

If you can't pick values of x that decide certain values, there is another option, although it's not as fun: find the powers of x on each side and equate them to get a system of equations that you can solve.
 
  • #7
LilTaru said:
Yeah. This was my first approach but if you plug in the value of a = 4, it seems the two factors of the hints never equal 0... Maybe this is just me.

They don't need to equal to zero necessarily, the equation would be true for all values of x.
 
  • #8
Okay... I tried to solve for the constants but I am stuck... I get this equation:

1 = (Ax + B)(x2 - 2*sqrt(2)*x + 4) + (Cx + D)(x2 + 2*sqrt(2)*x + 4)

I can't see how to solve this either by finding an x where the factors are zero or by solving a system of equations... How can I turn this into a system of equations?
 
  • #9
LilTaru said:
Okay... I tried to solve for the constants but I am stuck... I get this equation:

1 = (Ax + B)(x2 - 2*sqrt(2)*x + 4) + (Cx + D)(x2 + 2*sqrt(2)*x + 4)

I can't see how to solve this either by finding an x where the factors are zero or by solving a system of equations... How can I turn this into a system of equations?

Right, well you can multiply it out and equate coefficients OR you can put in values for x and get equations. For example, you can choose x=2,3,etc. and you will get equations to solve.
 
  • #10
But I thought when you put in values of x it should make one of the factors be 0 so that you can solve for the other one?
 
  • #11
LilTaru said:
But I thought when you put in values of x it should make one of the factors be 0 so that you can solve for the other one?

I did not mean it like that.

You have 4 unknowns and the equation is true for all values of x.

So you can put in 4 different values for x and get 4 equations.
 
  • #12
I've tried solving the system of equations but cannot seem to get it not matter how hard I try... can someone please help me solve the system? After that I know how to do the integral.
 
  • #13
What equations did you get?
 
  • #14
Main equation: 1 = (Ax + B)(x2 - 2[tex]\sqrt{}2[/tex]*x + 4) + (Cx + D)(x2 + 2[tex]\sqrt{}2[/tex]*x + 4)

Set x = 0: 1/4 = B + D

Set x = sqrt(8): 1 = 8*sqrt(2)A + 8B + 40*sqrt(2)C + 20D

Set x = sqrt(2): 1 = -2*sqrt(2)A - 2B + 28*sqrt(2)C + 14D

Set x = 3*sqrt(2): 1 = 30*sqrt(2) + 10B + 102*sqrt(2) + 34D

Which by looking at the answers seems impossible plus possibly horribly wrong. Can someone tell me what they get please?!
 
  • #15
After looking on the internet for a sample problem like this... I have found I am retarded... Figured out the problem: equated the coefficients wrong! Thank you all for your help! :P
 

1. What is the purpose of using a specific hint to evaluate an integral?

Using a specific hint can make the process of evaluating an integral more efficient and less time-consuming. It can also provide insight into the structure and behavior of the integral, making it easier to determine the best approach for solving it.

2. How do I choose the right hint for evaluating an integral?

Choosing the right hint depends on the characteristics of the integral, such as the form of the integrand and the limits of integration. Some common hints include substitution, integration by parts, and trigonometric identities.

3. Can using a specific hint always guarantee a solution to the integral?

No, using a specific hint does not always guarantee a solution to the integral. It is possible that the chosen hint may not lead to a closed form solution or may require additional techniques to fully evaluate the integral.

4. Are there any drawbacks to using a specific hint for evaluating an integral?

One potential drawback is that relying on a specific hint may limit the understanding of the integral and the overall problem solving skills. It is important to also have a strong foundation in basic integration techniques.

5. Can I use multiple hints to evaluate an integral?

Yes, it is possible to use multiple hints to evaluate an integral. In fact, some integrals may require a combination of techniques to successfully solve them. It is important to be flexible and use the hint(s) that seem most suitable for the given integral.

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