Page 12 - Integration Problems

1 = ( − 3) +

Expand

1 = − 3 +

Equate coefficients,

1 1
1 = −3 and 0 = + , = − and =
3 3

Therefore,

1 1 1
= − +
( − 3) 3 3( − 3)

The integration becomes,

2 ln( )
∫ 2
1 ( − 3)
2 2 2
ln( ) 1 1
= − [ ] + ∫ − ∫
( − 3) 3 ( − 3) 3
1 1 1

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

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