13.V(x-1)(x-2)

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Solution

The given function is 1 ( x−1 )( x−2 ) .

( x−1 )( x−2 )can be written as x 2 −3x+2 .(1)

Put (1) in the given function,

1 x 2 −3x+2 = 1 x 2 −3x+ 9 4 − 9 4 +2 = 1 ( x− 3 2 ) 2 − 1 4 = 1 ( x− 3 2 ) 2 − ( 1 2 ) 2 (2)

Also, ∫ 1 x 2 − a 2 = log| x+ x 2 − a 2 |+c (3)

Now let, x− 3 2 =t dx=dt

Substitute values of t and dt in (2),

1 ( x− 3 2 ) 2 − ( 1 2 ) 2 = 1 t 2 − ( 1 2 ) 2 =log| t+ t 2 − ( 1 2 ) 2 | =log| ( x− 3 2 )+ ( x− 3 2 ) 2 − ( 1 2 ) 2 | =log| ( x− 3 2 )+ x 2 −3x+2 | By Using (2)

Thus, the integral of the function 1 ( x−1 )( x−2 ) is log| ( x− 3 2 )+ x 2 −3x+2 |.

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