Integral of N^-1Date: 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 down. 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 Check out our web site! http://mathforum.org/dr.math/ |
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