Integration by PartsDate: 05/02/99 at 16:03:55 From: Steve Brown Subject: Integration by Parts Recently on an study exam for the AP Calc AB test, I ran across the following exercise: the general integral of x*ln(x)-1. I've determined that I need to do it by parts, since substitution fails. Date: 05/03/99 at 22:08:32 From: Doctor Jeremiah Subject: Re: Integration by Parts Hi Steve: I think you are right. Substitution did not work for me either. So we will do it by parts. The first thing to do is to simplify the integral. We need to break it up so that it will be easier to solve. / / / | x*ln(x) - 1 dx = | x*ln(x)*dx - | dx / / / In order to integrate by parts you need to break the integral into two pieces called "u" and "dv" . For the integral of x*ln(x)*dx we will choose u = ln(x) and dv = x dx . / / / | x*ln(x) - 1 dx = | u * dv - | dx / / / But we will also need to find "du" and "v" which can be found like this: u = ln(x) ==> du = 1/x dx dv = x dx ==> v = 1/2 * x^2 When you integrate by parts you follow the rule that says the integral of v * du is u*v minus the integral of v*du: / / | u * dv = u * v - | v * du / / So we will substitute in u*v minus the integral of v*du where we had the integral of v * du: / / / | x*ln(x)-1 dx = u * v - | v * du - | dx / / / / / / | x*ln(x)-1 dx = ln(x) * (1/2 * x^2)-|(1/2 * x^2) * (1/x * dx)-|dx / / / / / / | x*ln(x)-1 dx = ln(x) * (1/2 * x^2) - 1/2 * | x dx - | dx / / / And now we can solve the integrals: / | x*ln(x)-1 dx = ln(x) * (1/2 * x^2) - 1/2 * (1/2 * x^2) - (x) / Now to test our work. If we differentiate ln(x) * (1/2 * x^2) - 1/2 * (1/2 * x^2) - (x) with respect to x we get: d/dx( ln(x) * (1/2 * x^2) ) - 1/4 * d/dx(x^2) - d/dx(x) (1/2 * x^2) * d/dx( ln(x) ) + ln(x) * d/dx( 1/2 * x^2 ) - 1/4 * 2x - 1 (1/2 * x^2) * 1/x + ln(x) * 1/2 * d/dx( x^2 ) - 1/4 * 2x - 1 1/2 * x + ln(x) * x - 1/2 * x - 1 ln(x) * x - 1 Which is what we started with. Let me know if you need more details on how I solved it. - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ |
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