28.10 #20, from Basic Technical Mathematics with Calculus (8th Edition) by Allyn J. Washington.
                                                                   Z     3    2
                                                                     −x +x +x+3
                                                                                2      2 dx
                                                                      (x+1)(x +1)
                      This example involves a repeated irreducible quadratic factor. The partial fractions set-up is
                                                       3     2
                                                    −x +x +x+3 = A +Bx+C+ Dx+E .
                                                               2     2                2            2     2
                                                    (x+1)(x +1)            x+1       x +1       (x +1)
                                                             2      2
                      Multiplying both sides by (x + 1)(x +1) we obtain
                                        3    2                 2      2                       2
                                    −x +x +x+3=A(x +1) +(Bx+C)(x+1)(x +1)+(Dx+E)(x+1).                                                (1)
                                                  For x = −1, we get:        4=4A+0+0 =⇒ A=1.
                      To obtain B, C, D, and E we will expand equation (1) and equate coefficients of powers of x.
                          3     2                 4      2                     3    2
                       −x +x +x+3=A(x +2x +1)+(Bx+C)(x +x +x+1)+(Dx+E)(x+1)
                                                       4              3                        2
                                           =(A+B)x +(B+C)x +(2A+B+C+D)x +(B+C+D+E)x+(A+C+E)
                      By equating the corresponding coefficients of powers of x, we obtain
                                                                               A+B=0
                                                                               B+C=−1
                                                                   2A+B+C+D=1
                                                                     B+C+D+E=1
                                                                          A+C+E=3.
                      Since A = 1, we can deduce B, C, D, and E as follows.
                                                                       A+B=0 =⇒ B=−1
                                                                     B+C=−1 =⇒ C=0
                                                                  A+C+E=3 =⇒ E=2
                                                             B+C+D+E=1 =⇒ D=0
                      Therefore,             Z                           Z                                 
                                                  3     2
                                               −x +x +x+3dx=                   1    − x +             2        dx.
                                                          2     2                       2           2     2
                                               (x+1)(x +1)                   x+1 x +1 (x +1)
                      Wecanusethesubstitution u = x2 +1 to get
                                                              Z    x          1      2
                                                                x2 +1 dx = 2 ln(x +1)+C.
                      Wecanusethetrigonometric substitution x = tanθ to deduce that
                                                          Z   2 2    2 dx = 2x       +tan−1x+C.
                                                            (x +1)           x +1
                      Wethenconclude that
                                      Z     3     2
                                         −x +x +x+3                             1     2            x          −1
                                                   2      2 dx = ln|x+1|−         ln(x +1)+ 2           +tan      x+C.
                                         (x+1)(x +1)                            2               x +1
                      Gilles Cazelais. Typeset with LAT X on February 18, 2008.
                                        E