2 dx 1 = 9 1 1 1 1 x = ln 9 3 9 3(x 1) 6.5 Integration of expressions with quadratic factor in the denominator – proper fraction. Example 2 2  (1) dx The integrand is a proper fraction and has a quadratic factor. It is written as 3x 7   A Bx C 2 3x 7   1)  (Bx  1) A(x2 C)(x (x 1)(x2 1)   (x 1)(x2 1)  Then (3x + 7) = A (x2 + 1) + (B x + C)(x 1) (2) – We can put x = 1 on both of (2) to get 3 (1) + 7 = A (1 + 1) + (B + C) (0) 10 = 2 A or A = 5 Putting x = 0, 7 = A C, C = -2 – Equating coefficient of x2, 0 = A + B or B = -A = -5 Then 2 3x 7  5x dx dx  Hence  dx 1 1 1 2   x x x ) 5 = 5 ln (x 1) – – 2 ln (x2 + 1) 2 tan-1 x + C – 6.6 Integrand is an improper fraction If the degree of x in the Numerator is the same or greater that that of x in the denominator the Algebraic Fraction is called an improper fraction. 59 (x) F If the integrand (x) f
2 dx = = 1 9 1 9 ln x 1 3(x 1) 1 1 9 3 6.5 Integration of expressions with quadratic factor in the denominator – proper fraction. Example 2 2  dx (1) The integrand is a proper fraction and has a quadratic factor. It is written as 3x  7 A Bx  C 2 3x  7 (x 1)(x2  1)  A(x2  1)  (Bx  C)(x 1) (x 1)(x2  1) Then (3x + 7) = A (x2 + 1) + (B x + C)(x – 1) (2) We can put x = 1 on both of (2) to get 3 (1) + 7 = A (1 + 1) + (B + C) (0) 10 = 2 A or A = 5 Putting x = 0, 7 = A – C, C = -2 Equating coefficient of x2, 0 = A + B or B = -A = -5 Then 2 Hence 3x  7 2  dx  dx x 1 5x x  1 dx x  1 = 5 ln (x – 1) – 5 2 ln (x2 + 1) – 2 tan-1 x + C 6.6 Integrand is an improper fraction If the degree of x in the Numerator is the same or greater that that of x in the denominator the Algebraic Fraction is called an improper fraction. 59 If the integrand f (x) F (x) )