Expert: Paul Klarreich - 6/29/2007

Question
Hello!
I just can't figure out where to begin with this problem...

It's Integral from 0 to inf. of (X^3)/((X^2)+1)^2
One of my ideas was to do U subst. but it wouldn't work because of the X^3 in the numerator. Then I thought I can divide everything by X^4. Then took the integral of that and got 1/4 X^2 + 1/4 X^4. If this is correct then I can go ahead and use the limits... That's what I got so far.
Thank you in advance!

Answer
Questioner:   Annie
Category:  Calculus
 
Subject:  Calculus 2: Improper Integrals
Question:  Hello!
I just can't figure out where to begin with this problem...

It's Integral from 0 to inf. of (X^3)/((X^2)+1)^2
One of my ideas was to do U subst. but it wouldn't work because of the X^3 in the numerator. Then I thought I can divide everything by X^4. Then took the integral of that and got 1/4 X^2 + 1/4 X^4. If this is correct then I can go ahead and use the limits... That's what I got so far.
Thank you in advance!
..............................
Hi, Annie,

For your integral:

{inf       x^3
|      -----------  dx
}0     (x^2 + 1)^2

before you can evaluate it you have to be sure that it converges.  I don't think it does.  For large values of x, the integrand is asymptotic to  1/x, and that integral (from anything to infinity) diverges.  

If you want to be more formal about the proof:

Do some algebra:

{inf       x^3
|     ---------------  dx
}0    x^4 + 2x^2 + 1

To prove:
      x^3          1
 -------------- > ---
 x^4 + 2x^2 + 1    2x
This is true if:

x^4 > 2x^2 + 1

x^4 - 2x^2  > 1

x^2(x^2 - 2) > 1

If  x >= sqrt(3), then  x^2 >= 3  and  x^2 - 2 >= 1, so the LHS is > 1

So the integrand is > 1/2x for all x > sqrt(3).  That means:

{inf       x^3
|     ---------------  dx  >
}0    x^4 + 2x^2 + 1

{inf             x^3
|          ---------------  dx  >
}sqrt(3)    x^4 + 2x^2 + 1


{inf         1
|          ----  dx
}sqrt(3)    2x

But this diverges also.