Date: 06/18/97 at 00:27:03 From: Kevin Subject: Integration Hi, 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. Thanks! Kevin
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! http://mathforum.org/dr.math/
Date: 06/18/97 at 09:33:20 From: Doctor Anthony Subject: Re: Integration Kevin, 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) x^2 The second term can be written: ------------ x^3*sqrt(x^3-1) Now put in sqrt(x^3-1) = u. So (1/2)(x^3-1)^(-1/2)*3x^2 dx = du x^2dx Thus: --------- = (2/3)du sqrt(x^3-1) 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)du ---------- = (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! http://mathforum.org/dr.math/
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