Page 22 - Integration Problems

The integral becomes,

1 2 1 2 1 2
∫ = − ∫ + ∫
2 1 ( + 2)( − 2) 8 1 + 2 8 1 − 2

1
2
[−ln(| + 2|) + ln(| − 2|)]
1
8
1 − 2 2
= [ln (| |)]
8 + 2 1

But = 2 − 3

1 2 − 3 − 2 2
= [ln (| |)]
8 2 − 3 + 2 1

2
1 2 − 5
= [ln (| |)]
8 2 − 1 1

1 4 − 5 2 − 5
= [ln (| |) − ln (| |)]
8 4 − 1 2 − 1

1 −1 −3
= [ln (| |) − ln (| |)]
8 −3 −1

1 1
= [ln ( ) − ln(3)]
8 3

1 1 1
[ln ( ) + ln ( )]
8 3 3

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