Page 16 - Integration Problems

2
2
2
+ 4 + 29 = ( + ) +
2
= ( + 2) + 25
The integral becomes,

2 2
∫ 2 = ∫ 2

0 + 4 + 29 0 ( + 2) + 25
Let = + 2, =

2 2
∫ 2 = ∫ 2 2

0 ( + 2) + 25 0 + 5
Using the well known result,

1
∫ = tan −1 ( )
2
+ 2

2 1 2
∫ 2 2 = [ tan −1 ( )]
0 + 5 5 5 0

But = + 2

2 1 + 2 2
∫ 2 = [ tan −1 ( )]
0 ( + 2) + 25 5 5 0

1 2 + 2 2
= [tan −1 ( ) − tan −1 ( )]
5 5 5

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