Page 13 - Integration Problems

ln(2) ln(1)
= − +
(2 − 3) (1 − 3)
1 |2 − 3| 1 |1 − 3|
+ ln ( ) − ln ( )
3 |2| 3 |1|

ln(2) ln(1) 1 1 1 2
= − + ln ( ) − ln ( )
1 2 3 2 3 1

1 1 ln(2)
= ln(2) − ln(2) − ln(2) =
3 3 3

2 ln( ) ln(2)
∫ 2 =
1 ( − 3) 3
The keystrokes for this calculation,

h2)a3=

The original integration is,

2 ln( )
∫ 2
1 ( − 3)

Verify the result by using the integration feature on

your fx calculator,

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