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Willow Pratt 2022-07-06 Answered
I have to evaluate this integral:
1 x + x 2 x 1 + x x 2 d x
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Answers (2)

esperoanow
Answered 2022-07-07 Author has 11 answers
You can split this into three integrals:
I 1 = 1 x 1 + x x 2 d x
I 2 = 1 1 + x x 2 d x
I 3 = x 1 + x x 2 d x
For I 1 , substitute t = 1 x and this leads to an acrosh-type integral.
For I 2 , this is an arcsin-type integral.
For I 3 , you can rewrite it as
1 2 2 x + 1 1 + x x 2 d x + 1 2 I 2
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Araceli Clay
Answered 2022-07-08 Author has 2 answers
1 x + x 2 x 1 + x x 2 d x
The rooted is a trinomial ( a x 2 + 2 b x + c), which has two real roots α and β, decomposing it we have: a x 2 + 2 b x + c = a ( x α ) ( x β ) = ( x β ) a ( x α ) x β . Let's say a ( x α ) x β = t 2 , and we get :
x = a α β t 2 a t 2
d x = 2 a t ( α β ) ( a t 2 ) 2 d t
Substituting these values in the integral, we have: 1 x + x 2 x 1 + x x 2 d x =
= 4 ( 2 t 4 t 2 + 2 ) ( t 2 + 1 ) 2 ( t 2 ( 5 + 1 ) 5 + 1 ) d t =
expanding
= 2 5 ( t 2 + 1 ) 2 d t 1 5 t 2 + 1 d t 4 t 2 ( 5 + 1 ) 5 + 1 ) d t
Calculating the individual integrals:
I 1 = 5 . t a n 1 ( t ) + 5 t t 2 + 1
I 2 = ( 1 5 ) . t a n 1 ( t )
I 3 = l n ( 2 t 5 + 1 2 t + 5 1 )
Ultimately you have: I = I 1 + I 2 + I 3 = t a n 1 ( t ) + 5 t t 2 + 1 l n ( 2 t 5 + 1 2 t + 5 1 )
All that remains is to replace the value of t with x:
t = 2 x 5 + 1 2 x 5 1
getting the result.
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