Integration by Partial Fraction Decomposition
Date: 06/24/99 at 05:20:12 From: Robert Subject: Integral problem I am having problems doing this kind of problem: integral of dx/(x^4+x^2) I tried factoring an x^2 out, but this doesn't seem to help. Also, integral of xdx/(x^2+2x+1) I tried doing this by parts, u= x, du = dx and dv = 1/((x+1)^2+1), v = arctan(x+1) but this doesn't seem to work either. HELP! Thanks, Robert
Date: 06/24/99 at 08:13:07 From: Doctor Jerry Subject: Re: Integral problem Hi Robert, What you need is "partial fraction decomposition." 1/(x^4+x^2) = 1/((x^2+1)(x^2)). Using partial fraction principles, we know that this fraction came from fractions of the form (A*x+B)/(x^2+1) + C/x + D/x^2, where A, B, C, and D are unknowns, to be determined. In this particular case, C = 0, D = 1, A = 0, and B = -1. That is, 1/(x^4+x^2) = 1/x^2 - 1/(1+x^2). Now the integration is easy. For the other, x/(x^2+2x+1) = x/(x+1)^2. Here the partial fraction decomposition must have the form A/(x+1) + B/(x+1)^2. In this case, A = 1 and B = -1. Now the integration is easy. How do you find A and B? Set A/(x+1) + B/(x+1)^2 = x/(x+1)^2 Clear fractions by multiplying by (x+1)^2: A(x+1)+B = x. This must hold for all x. Let x take on two nice values to obtain two equations in two unknowns. Here, take x = -1 and x = 0: B = -1 A+B = 0 So, B = -1 and A = 1. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/
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