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EXAMPLES OF CASE I

Example 1: Evaluate ∫ 2 x 2 -x +6 6 dx, (x > 2)
- 7x +

Solution: The denomicator 2x2 - 7x + 6 = (x - 2) (2x - 3),

Let ( x --2x)(+2x=6 - 3) A+ B
x - 2 2x - 3

or -x + 6 = A(2x - 3) + B(x - 2) which is true for all x
Putting x = 2, we get
-2 + 6 = A(4 - 3) + B x 0 ⇒ A = 4

and Putting =x 3 , we get - 3 + 6= A(0) + B 3 - 2 
22  2 

or 9 =B  - 1 ⇒B =- 9
2  2 

Thus ∫ ( x -x + 6 3) dx = ∫  x 4 2 + -9  dx
 - 2x - 3 
- 2)(2x -

=4 ∫( x - 2)-11 . dx - 9 ∫(2x - -1
2
3) . 2dx

= 4 ln ( x - 2) - 9 ln (2x - 3) + c, ( x > 2)
2

Example 2: ∫Evaluate 2x3 - 9x2 + 12x dx, (x > 2)
2x2 - 7x + 6

Solution: After performing the division by the denominator, we get

∫ ∫2x3 - 9x2 + 12xdx=  x - 1 + -x +6  dx
 2x2 - 7x + 6 
2x2 - 7x + 6

∫ ∫ ∫ ∫= 4 -9 dx,
x dx - 1 dx + (x - 2) dx + 2x - 3 (See the Example 1)

= x2 - x + 4 ln ( x - 2) - 9 (2x - 3) + c, ( x > 2)

22

version: 1.1

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