Page 61 - Integration Problems

The integral becomes,
1 1 1
1 1
∫ − ∫ + ∫
2 + 3 + 2 2 + 1
−0.5 −0.5 −0.5

1 1 1
= [ ln( + 3) − ln( + 2) + ln( + 1)]
2 2 −0.5
1 1
= [ ln{( + 3)( + 1)} − ln( + 2)]
2 −0.5

1
= [ ln{(1 + 3)(1 + 1)} − ln(1 + 2)]
2

1
− [ ln{(−0.5 + 3)(−0.5 + 1)} − ln(−0.5 + 2)]
2

1 1
= [ ln{(4)(2)} − ln(3)] − [ ln{(2.5)(0.5)} −
2 2
ln(1.5)]

1 1
= ln{8} − ln(3) − ln{1.25} + ln(1.5)
2 2

1 8 1.5
= ln ( ) + ln ( )
2 1.25 3
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