[TI-M] Re: An Integral, hehehe

To: ti-math@lists.ticalc.org
Subject: [TI-M] Re: An Integral, hehehe
From: JasonScho@aol.com
Date: Thu, 2 May 2002 08:59:09 EDT
Reply-To: ti-math@lists.ticalc.org


/  5+x^2
|  ----- dx
/  9-x^3

Let a^3=9 (to make it easier to follow)

9 - x^3 = (a - x)(a^2 + ax + x^2)

(5 + x^2) / (9 - x^3) = A / (a - x) + (Bx + C) / (a^2 + ax + x^2)

A(a^2 + ax + x^2) + (Bx + C)(a - x) = 5 + x^2

a=x -> 3Aa^2 = 5 + a^2
A = (5a + 3) / 9

Equating coefficients of x^2: A - B = 1
B = (5a - 6) / 9

Equating constant terms: Aa^2 + Ca = 5
aC = 5 - (5 + a^2) / 3 = (10 - a^2) / 3
C = (10a^2 - 3a) / 9

(5+x^2)/(9-x^3) = (5a - 3) / 9(a - x) + ((5a - 6)x + (10a^2 - 3a)) / 9(a^2 + 
ax + x^2)
   = (5a - 3) / 9(a - x) + (2x + a)(2.5a - 3) / 9(a^2 + ax + x^2) + 7.5a / 
9((x + a/2)^2 + (3/4)a^2)

Integrating each part gives:

(1/9) * ((3 - 5a) ln abs (a - x) + (2.5a - 3) ln (a^2 + ax + x^2) + 5a 
sqrt(3) / a * arctan((2x + a) / (a sqrt(3)))

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