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Choosing Factors When Integrating by Parts

Date: 10/08/2003 at 21:23:52
From: Consuelo  
Subject: Integration by parts

I use integration by parts to find 

   1
  / 
  | (r^3)/[(4+r^2)^(1/2)] dr
  /
  0

What I find most difficult is finding u, v, du, and dv.

I tried separating the integral by making one simple integral to take 
the anti-derivative of and then using the difficult integral to do 
integration by parts.  I just don't know how to start it.


Date: 10/11/2003 at 13:53:48
From: Doctor Fenton
Subject: Re: Integration by parts

Hi Consuelo,

Thanks for writing to Dr. Math.  I like to write the integration
by parts formula in the form

   /                /
   | u v' dx = uv - | u' v dx ,
   /                /

because it arises from the Product Rule for derivatives:

   [u v]' = u' v + u v' .

The idea is to look at the integrand as a product of two functions,
and one of them must be an exact derivative (or maybe a constant
times an exact derivative).  The trick is to choose the factors
correctly.  

Formally, it looks as if you are "moving" the derivative from one
factor to the other.  This will help only if the new integrand is
simpler in some sense than the original.

In your example, the integrand is 

  (r^3)/[(4+r^2)^(1/2)]

The square root is a unit which you can't break up, so the only
question is how to distribute the powers of r.  Essentially, the
only choices are to take the factors to be one of the following:

     r^3   and       (4+r^2)^(-1/2)  

     r^2   and     r*(4+r^2)^(-1/2)  

     r     and   r^2*(4+r^2)^(-1/2)  

Of course, we also have to choose one to be u (the factor to be
differentiated) and one to be v' (the factor to be integrated).

You can differentiate 

  (4+r^2)^(-1/2)

but if you do, you will obtain a factor of 

  (4+r^2)^(-3/2)

plus an additional power of r from applying the Chain Rule.  You would
also have to integrate r^3, which would produce r^4, so the integrand
in the new integral after using this approach would be a constant times 

  r^5*(4+r^2)^(-3/2)

That looks worse than the original integral: it has more powers of r,
and a larger power of (4+r^2) in the denominator. This observation 
strongly suggests that you want to integrate 

  (4+r^2)^(-1/2).

However, if you try to use the substitution 

  u = 4+r^2

then du = 2rdr, so you need a power of r to integrate this term.  That
means that you should write the original integrand as the product of
r^2 and 

  r*(4+r^2)^(-1/2)

taking

  u = r^2  and v' = r*(4+r^2)^(-1/2) .

After integrating, you will have an integrand which is a constant 
times 

  r*(4+r^2)^(1/2)

and that can be evaluated with another integration by substitution.

If you have any questions or need more help, please write back and
show me what you have been able to do, and I will try to offer further
suggestions.

- Doctor Fenton, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Calculus
High School Calculus

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