Integrate by Partial Fractions Use the method of partial fractions to compute the integral: � x2 + 2x + 3 dx. (x + 1)(x + 2)(x + 3) Solution We ﬁrst check to see if we can factor the numerator to cancel any terms in the denominator; we can’t. Since all the terms in the denominator are linear, we need not try to factor them. The numerator is a second degree polynomial and the denominator is third degree, so we do not need to perform any long division of polynomials. We set up the partial fractions decomposition as follows: A B C x2 + 2x + 3 = + + . (x + 1)(x + 2)(x + 3) x+1 x+2 x+3 Now we apply the cover-up method, starting by solving for A: � A ��B��C� x2 + 2x + 3 = �� + �� +�� . � � � � � � (x +� 1)(x + 2)(x + 3) � x +� 1�x + 2 x +� 3 � � We cover up x + 1 and any expression that does not contain x + 1, then plug in x = −1; the value that makes x + 1 equal 0. We get: (−1)2 + 2 · (−1) + 3 = 1 = A. (−1 + 2)(−1 + 3) We repeat this process for B and C: x2 + 2x + 3 � (x + 1)� (x� +� 2)(x + 3) � � � (−2)2 + 2(−2) + 3 (−2 + 1)(−2 + 3) B = �A �� B � C�� +� �+ � � �� x� +� 1�� x+ 2�� x+ 3 � � = B = −3 x2 + 2x + 3 � (x + 1)(x + 2)� (x� +� 3) � � � 2 (−3) + 2(−3) + 3 −3 + 2 (−3 + 1 C = �A � �� B�� + C �+ � � � � � x + 1 x + 2�� x+ 3 � � = C = 3 We conclude that: x2 + 2x + 3 1 3 3 = − + . (x + 1)(x + 2)(x + 3) x+1 x+2 x+3 1 If we plug in x = 0 we get 12 = 1 − 23 + 1, so this is probably a correct decom position. We can now break this down into three relatively simple integrals. One integration is presented in detail below, using the substitution u = x + 3; the other two are similar: � � 3 1 dx = 3 du x+3 u = 3 ln |u| + c = 3 ln |x + 3| + c In conclusion, � x2 + 2x + 3 dx = (x + 1)(x + 2)(x + 3) = � 1 3 3 − + dx x+1 x+2 x+3 ln |x + 1| − 3 ln |x + 2| + 3 ln |x + 3| + c We could try to check this by diﬀerentiation, but that leads directly to verifying that our partial fractions decomposition is correct — a time consuming task.