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Integral of N^-1

Date: 09/26/97 at 18:36:33
From: Mimi
Subject: Differential calculus

How do I find the integral of a number raised to the -1st power?
For example:
       {(3x^3)+5/x)^-1 dx

Date: 09/27/97 at 06:39:24
From: Doctor Mitteldorf
Subject: Re: Differential calculus

Dear Mimi,

Integrals like this are not usually easy, and most of them can't be 
"done" at all.  It just means that the integral is some function that 
doesn't have a name and there's no other way to express it in terms of
elementary operations like addition, multiplication, squaring, etc.  
The integral still exists - it's just not something you can write 

This one, however, happens to be more tractable. First of all, 
remember that something to the -1 power is just the reciprocal of that 
something. Then multiply top and bottom by x to "clear" the fraction.  
You get

   Integral of x dx / (3x^4 + 5)

Next, use the substitution u = x^2.  I got that idea from noticing 
that the numerator xdx is an exact differential of x^2/2, and the 
denominator has only the square of x^2 in it, with no leftover x's.  
So, in terms of u,

   1/2 Integral of  du / (3u^2+5)

The next idea, this time coming from experience doing integrals 
and from the trig identity 1 + tan^2 = sec^2, is to substitute 
3u^2 = 5 tan^2(t), which is the same as u = 5/3 tan(t). This leads 
to du = 5/3 sec^2(t) dt. In terms of t, the integral is

   1/2 Integral 5/3 sec^2(t) dt / (5tan^2(t)+5)
   1/2 Integral 5/3 sec^2(t) dt / (5sec^2(t))
   1/6 Integral dt 

This is just t/6 + const.

Now you just go back and write t in terms of u:  t = arctan(3u/5).
Then write u in terms of x: u = x^2.

The answer, then, is that the integral is 1/6 arctan(3x^2/5).

Congratulations if you've followed all this. It's pretty advanced for 
a 16-year-old. The idea of substitutions in an integral takes some 
getting used to.  Please try some more examples, and write back as you 
develop your facility with the techniques.

-Doctor Mitteldorf,  The Math Forum
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Associated Topics:
High School Calculus

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