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Date: 06/18/97 at 00:27:03
From: Kevin
Subject: Integration 


I have this integral that seems impossible. There seems to be nothing 
in my tables that I can use and I've tried to rearrange the problem to 
make it work, but with no luck. The question is the unbounded integral 
(anti-derivative) of:

sqrt(x^3-1)/x dx

I rewrote the problem as integral of 1/x*(x^3-1)^1/2, but it really 
didn't help. Integration by parts didn't seem to help either since 
d(x^3-1)^1/2 is like 1/2(x^2-1)(3x^2), which doesn't seem to help me 
at all.  I'm really stuck. This thing is tough. I'd really appreciate
any advice you can offer.  


Date: 06/18/97 at 08:29:22
From: Doctor Jerry
Subject: Re: Integration 

Hi Kevin,

Try the substitution x^3-1 = w.  From 3x^2 dx = dw, you can see that
dx/x = dw/(3(w+1)).

You will obtain an integral for which the substitution w = tan^2(t) 
works out well.

-Doctor Jerry,  The Math Forum
 Check out our web site!   

Date: 06/18/97 at 09:33:20
From: Doctor Anthony
Subject: Re: Integration 


Start by multiplying the top and bottom by sqrt(x^3-1).

               x^3 -1             x^2               1
This gives:  ----------    =  ----------    -  ------------
            x*sqrt(x^3-1)     sqrt(x^3-1)      x*sqrt(x^3-1)

The first term is easy. Put in x^3-1 = u, 3x^2dx = du, x^2dx = 1/3du.

So the first term is 1/3du/u^(1/2), which integrates to:

2/3u^(1/2) = (2/3)sqrt(x^3-1)

The second term can be written:   ------------

Now put in sqrt(x^3-1) = u. So (1/2)(x^3-1)^(-1/2)*3x^2 dx = du  

Thus:  ---------   = (2/3)du

Also if we square sqrt(x^3-1) = u  we get:  x^3-1 = u^2
                                              x^3 = u^2+1

The whole of the second expression can therefore be written:

      ----------  =  (2/3)tan^(-1)(u) =  (2/3)tan^(-1)sqrt(x^3-1)
       u^2 + 1

The complete integral is therefore:

    (2/3)sqrt(x^3-1) - (2/3)tan^(-1)sqrt(x^3-1)  +  constant.

-Doctor Anthony,  The Math Forum
 Check out our web site!   
Associated Topics:
High School Calculus

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