Suppose C1 and C2 are physically the same curve, but they are parameterized so that the starting point of C1 is the ending point of C2, and
the ending point of C1 is the starting point of C2.
Express
∫ C1m(x, y)D „ x + n(x, y)D „ y
in terms of
∫ C2m(x, y)D „ x + n(x, y)D „ y.
∫ C1m(x, y)D ‚x + n(x, y)D ‚y = -∫ C2 m(x, y)D ‚x + n(x, y)D ‚y.

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(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0