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Integral calculus
  Integrating rational functions
  Recall that a rational function is a ratio of two polynomials written  Pn(x) / Qm(x).
To integrate a rational function
in case n > m, that is, if the degree of P is greater than or equal to degree of Q, we first divide Pn(x) by Qm(x) to obtain the sum of the polynomial Pn -  m(x) and a proper or simple rational function.
Then, we integrate proper rational function using decomposition of rational function into a sum of partial fractions.
  Use of the partial fraction decomposition to integrate a proper rational function
Any polynomial with real coefficients can be factored uniquely into a product of its lading coefficient, linear factors of the form (x - c) that correspond to its real roots, and irreducible quadratic polynomials that correspond to pairs of conjugate complex roots.
Therefore, to represent a proper rational function as a finite sum of terms we factor the polynomial Q in the denominator into powers of distinct linear terms and/or powers of distinct irreducible quadratic polynomials, i.e., which do not have real roots. Thus,
 - if the factors in the denominator Q are of the form (x - c) then the partial fractions will be of the form
where c is a real root of order k and A1, A2, . . . , Ak are unknown constants.
 - If the factors of Q are irreducible quadratic polynomials
of order k then, the partial fractions will be of the form
where M1, M2, . . . , Mk  and  N1, N2, . . . , Nk  are unknown constants.
Note that the leading coefficient a and the vertical translation y0 of a quadratic polynomial should be of the same sign the quadratic to have pair of conjugate complex roots a bi.
 Use of the partial fraction decomposition to integrate a proper rational function examples
 Example:  Evaluate
 Solution:

 Example:  Evaluate
 Solution:
Unknown constants  A, B, and C can be calculated by plugging suitable values for x, like roots for example, into given equation. Thus, by plugging the following values for x into the equation
x2 - x + 3 = A (x - 1)(x + 2) + B (x + 2) + C (x - 1)2
    for      x = 1       obtained is    3 = 3B,                       B = 1
    for      x = - 2    obtained is    9 = 9C,                      C = 1
    for      x = 0       obtained is    3 = - 2A + 2B + C,   and after substituting  B = 1 and  C = 1
                                   we get    3 = - 2A + 2 + 1,       A = 0.

 Example:  Evaluate
 Solution:
 
 
 
 
 
 
 
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