Antiderivative of the following function

C, right?Yes, that is correct. Remember to apply the chain rule when integrating the second integral. The derivative of tan^-1(x/2) would be 1/(4+x^2) * 1/2. So the final answer would be 2tan^-1(x/2) - 1/2 * 1/(4+x^2) + CIn summary, the method for finding the antiderivative of the function 8u^5 / ((4 + u^4)^2) involves rearranging the equation and using substitution, followed by clever algebra and trigonometric substitution. The final solution is 2tan^-1(u^2/2) - 1/2
  • #1
KataKoniK
1,347
0
How would you guys come about getting the antiderivative for the following function?

8u^5 / ((4 + u^4)^2)

I tried rearranging the equation like the following:

8u^5 * ((4 + u^4)^(-2)), but it's not really getting me anywhere. I tried to do a u substitution, but can't find one where it knocks off the top u^5.

Thanks in advance.
 
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  • #2
I take it that you're taking the A.D. wrt u?
My TI89 gives:
pi*atan(u^2/2)/180 - (2*u^2/(u^4+4))
Which in turn tells me that your calculus text should have a table for it in one of the appendices.
Look for it in the TOC, or index.
Look for an indefinite integral that has the general form:
a*x^(n)/(x^(n-1) +(n-1))^(2)
Or something similar.
 
  • #3
Integration by parts and you'll have to use trig substitution. That's a nasty one. Good luck!
 
  • #4
Here are a couple of hints. Try the substitution x = u2. That should help reduce things to a more manageable integrand. After that, a little clever algebra and a trig substitution will get you there. Not a trivial problem.

hotvette
 
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  • #5
hotvette said:
Here are a couple of hints. Try the substitution x = u2. That should help reduce things to a more manageable integrand. After that, a little clever algebra and a trig substitution will get you there. Not a trivial problem.

hotvette


Hmm, any hint on what the clever algebra should be? I can't really figure it out from here. I have simplified the equation down to

4x2 / (4 + x2)2 dx

using x = u2
 
  • #6
Try adding and subtracting the same number from the top. Gather terms and split the problem into 2 problems.

Example: if I have y in the numerator, y - 2 + 2, or (y - 2) + 2 is the same thing. Just pick the right number to add and subtract.
 
  • #7
Thanks for the reply. Am I on the right track?


4x2 / (4 + x2)2 dx
= (4x2 + 16 - 16) / (4 + x2)2 dx
= 4((x2 + 4) - 4)) / (4 + x2)2 dx
= (4 / (4 + x2)) - (4/ (4 + x2)2) dx
 
  • #8
You bet. The first one is readily integrated and the second is w/ a trig substitution. My college calc book has an entire section on appropriate trig substitutions based on what the integrand looks like - that's where I found the appropriate trig substituion.
 
  • #9
Thanks. However, can you spot my mistake when solving for the antiderivative of (4/ (4 + x2)2)? I must be doing something wrong, but I can't find what

(4/ (4 + x2)2)

Let x = 2 tan theta
dx = 2 sec2theta dtheta

= (8 sec2theta / ((4 + 4tan2theta)2)

factor out a 2

= 4 + 4 tan2 theta / ((4 + 4tan2theta)2)
= 1 / (4 + 4tan2theta)
= tan inverse (tan theta)
 
  • #10
The part I don't follow is how you get from:

[tex]\frac {d \Theta}{(4 + 4 tan^2 \Theta)}[/tex]

to

[tex]tan^{-1} tan \Theta[/tex]

By the way, I had a slightly different integrand, with 16 in the numerator instead of 4, but frankly, I don't want to take the time to re-check it. Mine might be wrong, I don't know. My objective here is getting you going on the right path, not necessarily verifying the exact answer.
 
  • #11
Alright, thanks. I'll just try it again.
 
  • #12
KataKoniK said:
Hmm, any hint on what the clever algebra should be? I can't really figure it out from here. I have simplified the equation down to

4x2 / (4 + x2)2 dx

using x = u2

I think there's a problem in this substitution.

[tex]\int \frac{8u^5}{(4+u^4)^2}du[/tex]

[tex]u=x^2[/tex]
[tex]du=2xdx[/tex]
 
  • #13
Jameson said:
I think there's a problem in this substitution.

[tex]\int \frac{8u^5}{(4+u^4)^2}du[/tex]

[tex]u=x^2[/tex]
[tex]du=2xdx[/tex]

No. The substitution is [itex]x=u^2[/itex]. You have it backwards.

[tex]\int \frac{4x^2}{(4+x^2)^2} \ dx[/tex]

is correct after the substitution.
 
  • #14
I just checked my work again. Here's what I get:

[tex]\int \frac{4x^2}{(4+x^2)^2} \ dx \ = \ \int \frac{4}{(4 + x^2)} \ dx \ - \int \frac{16}{(4 + x^2)^2} \ dx [/tex]

Btw, it might be prudent to reduce even further by factoring out the 4's to make it (1 + something^2) instead of (4 + x^2). May not be critical, though.
 
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  • #15
Yup, that's what I got after the substitution and working. Sorry for replying so late. I finally had a chance to do this question just now.


[tex] \int \frac{16}{(4 + x^2)^2} \ dx = \frac{32sec^{2}\Theta}{(4 + 4tan^{2}\Theta)^2} [/tex] d theta

where [tex] x = 2tan\Theta[/tex]
[tex] \ dx = 2 sec^{2}\Theta[/tex] d theta

After all that, the result I get is [tex]\Theta + \frac{sin2\Theta}{2}[/tex], but for some reason, when I use my calculator to see if this answer is correct, it doesn't match up with the derivative values of the original equation 8u^5 / (4 + u^4)^2
 
  • #16
Don't forget the first integral.
 
  • #17
Yup, I checked it with both the first and the second integral and no dice. The antiderivative for

[tex]\int \frac{4}{(4 + x^2)} \ dx[/tex]

is [tex]2tan^{-1}(x/2)[/tex]
 

What is an antiderivative?

An antiderivative, also known as an indefinite integral, is the inverse operation of a derivative. It is a function that, when differentiated, produces the original function.

How do you find the antiderivative of a function?

To find the antiderivative of a function, you can use integration techniques such as u-substitution, integration by parts, or the fundamental theorem of calculus. The result will be a family of functions, as there are multiple possible antiderivatives for a single function.

Can all functions have an antiderivative?

No, not all functions have an antiderivative. Functions that are not continuous or have "breaks" in their graph, such as the absolute value function, do not have an antiderivative. These functions are said to be non-integrable.

What is the difference between a definite and indefinite integral?

A definite integral has specific limits of integration and gives a numerical value, while an indefinite integral does not have limits and results in a family of functions. The definite integral can be thought of as finding the area under a curve, while the indefinite integral is finding a function that when differentiated, gives the original function.

Why is the antiderivative important in calculus?

The antiderivative is important in calculus because it allows us to find the original function from its derivative, which has many real-world applications. It also plays a key role in finding areas and volumes using integration, and is a fundamental concept in understanding the behavior of functions.

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