is an improper fraction we shall divide f (x) by F (x) and the remainder is resolved into partial fractions. This principle is illustrated from the following example: Example 3 Find 3x2 2x 7 dx 2 x2 x Solution The integrand is an improper algebraic fraction. using long division we find that the quotient is 3 and the remainder is (x 1) – 3x2 2x 7 Then (x 1) 2 (x 2)(x  1) 3   Note: sometimes instead of long division it is easier to proceed as follows: 3x2 2x 7 3(x2 2) 1 2  x x x2 x  (x2 2) x 1 x = 3  2 (x 1) = 3  (x 2)(x 1)  1 x Now we can resolve (x 2)(x  1) into partial fractions. 1 x A B Let   (x 2)(x 1) 2  1  x x  1)  2) A(x B(x = (x 2)(x 1)  (x 1) = A (x + 1) + B (x 2) (2) – – 1 Putting x = 2 on both sides of (2) A = 3 . 3 Putting x = -1 on both sides of (2), we have 2 = B (-3) or – 2 =B  x h ) 1 e  (x 2)(x
is an improper fraction we shall divide f (x) by F (x) and the remainder is resolved into partial fractions. This principle is illustrated from the following example: Example 3 Find 3x2 2x 7 x2 x 2 dx Solution The integrand is an improper algebraic fraction. using long division we find that the quotient is 3 and the remainder is (x – 1) Then 3x2 2x 7 2  3  (x 1) (x 2)(x  1) Note: sometimes instead of long division it is easier to proceed as follows: 3x2 2x 7 x2 x 2  3(x2 x 2)  x 1 (x2 x 2) = 3  = 3  x 1 2 (x 1) (x 2)(x  1) Now we can resolve x 1 (x 2)(x  1) into partial fractions. Let x 1 (x 2)(x  1)  A x 2  B x  1 = A(x  1)  B(x 2) (x 2)(x  1) (x – 1) = A (x + 1) + B (x – 2) (2) Putting x = 2 on both sides of (2) A = 1 3 . Putting x = -1 on both sides of (2), we have – 2 = B (-3) or 2 3 =B x 1 (x 2)(x   ) e h
CALCULUS%20II.DOCX