problems with integration

[resolved]

using the Double Integration Method to determine the deflection in a beam. I need to integrate the below equation twice.

y2= 168 (-8 + x)^2/2 + 10 (-8 + x)^3/3 - 4 (-6 + x)^2/2 + 5 (-4 + x)^3/3 + 96 x^2/2 - 15 x^3/3

by hand I get:

y = 28 (-8 + x)^3 + 5/6 (-8 + x)^4 - 2/3 (-6 + x)^3 + 5/12 (-4 + x)^4 + 16 x^3 - (5 x^4)/4

when I integrate y2 in the HP Prime I get

y = -5/4*x^4 + 5/12*(x-4)^4 + 5/6*(x-8)^4 + 16*x^3 - 4*(1/6*x^3 - 3*x^2) + 168*(1/6*x^3 - 4*x^3)

HP Prime integrates the first four terms using the power formula and then switches in the last two terms by expanding the terms and integrating by differences and then simplifying.

-4*(x-6) integrates to -2*(x-6)^2 which integrate to -2/3*(x-6)^3

-2/3*(x-6)^3 Expanded = -2/3*x^3 + 12*x^2 - 72*x + 144

where HP expands first, -4x+24, which integrates to -2x + 24x which integrate to -2/3*x^3 + 12 x^2

clearly both are correct mathematically, but I have lost a lot of information by the way HP Prime integrated the last two terms. Why does HP change its methodology half way through integration?? is there any way to force use the power formula over the whole equation so I don't lost this information.

when solving for c1x and c2, the c2 term drops out x=0, y=0, c2=0 , but I get different values for c1 when evaluating the beam at its end x=10, y=0 to determine the value of c1, as would be expected between the two different equations. by hand c1 = -6352/15 , but HP evaluates to +10384/3. Clearly, my instructor would not give my credit for an incorrect answer if I told him the difference was in the c2 term.