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Crystal Wheeler 2022-07-02 Answered
Integrate:
x 2 x + 1 d x
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Answers (2)

Alexis Fields
Answered 2022-07-03 Author has 14 answers
I = x 2 x + 1 d x = x x 2 x + 1 x 2 x 1 2 x 2 x + 1 d x = x x 2 x + 1 2 ( x 2 x + 1 ) + ( x 1 2 ) 3 2 2 x 2 x + 1 d x = x x 2 x + 1 I 1 4 2 x 1 x 2 x + 1 d x + 3 4 d x x 2 x + 1 = 1 2 ( x x 2 x + 1 1 2 x 2 x + 1 ) + 3 8 d x ( x 1 2 ) 2 + 3 4
For the last integral, letting x 1 2 = 3 2 tan θ yields
d x ( x 1 2 ) 2 + 3 4 = 3 2 sec 2 θ d θ 3 4 tan 2 θ + 3 4 = sec θ d θ = ln | sec θ + tan θ | + C 1 = ln | 2 x 2 x + 1 3 + 2 x 1 3 | + C 1 = ln | 2 x 1 + 2 x 2 x + 1 | + C 2
Now we can conclude that
I = 2 x 1 4 x 2 x + 1 + 3 8 ln | 2 x 1 + 2 x 2 x + 1 | + C

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Lillianna Andersen
Answered 2022-07-04 Author has 3 answers
Alternate Method (Using hyperbolic functions)
Letting x 1 2 = 3 2 sinh θ yields
d x ( x 1 2 ) 2 + 3 4 = 3 2 cosh θ d θ 3 4 sinh 2 θ + 3 4 = d θ = θ = sinh 1 ( 2 x 1 3 )
I = 2 x 1 4 x 2 x + 1 + 3 8 sinh 1 ( 2 x 1 3 ) + C

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