Page 50 - Integration Problems
P. 50
The result is,
2 1
= − 1 +
+ 1 + 1
The integral becomes
3 3
1 2 1 1
∫ = ∫ [ − 1 + ]
2 + 1 2 + 1
0 0
Performing the integration,
1 2 3
= [ − + ln(| + 1|)]
2 2
0
The complete integral is,
3 3
2
∫ ln(1 + ) = [ ln(1 + )]
2
0 0
1 2 3
− [ − + ln(| + 1|)]
2 2
0
2 2 1 3
= [ ln(1 + ) − + − ln(| + 1|)]
2 4 2 2
0
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