Solve the integral int(1+x+sqrt(1+x^2))(sqrt(x+1)+sqrt(x))dx

propappeale00 2022-10-16 Answered
Solve the integral
1 + x + 1 + x 2 x + 1 + x d x
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Answers (2)

Plutbantonavv
Answered 2022-10-17 Author has 15 answers
One can be brought to the form u a ( 1 u ) b d u which is discussed in this answer of mine:
J = x 1 + x 2 d x = u = x 2 1 2 u 1 / 4 ( 1 + u ) 1 / 2 d u = 2 3 u 3 / 4 2 F 1 ( 1 2 , 3 4 ; 7 4 ; u ) + const. = 2 3 x 3 / 2 2 F 1 ( 1 2 , 3 4 ; 7 4 ; x 2 ) + const.
The other integral is an elliptic integral:
I = 4 15 2 α ( α F ( arcsin ( x 1 α x + α ) , α 2 ) 1 α E ( arcsin ( x 1 α x + α ) , α 2 ) ) + 2 15 1 + x 1 + x 2 ( 3 x + 1 + 4 x + α ) + const.
where α = 2 + 1 and
E ( ϕ , m ) = 0 ϕ 1 m sin 2 φ d φ , F ( ϕ , m ) = 0 ϕ d φ 1 m sin 2 φ
It can be evaluated using the Jacobi elliptic functions substitution:
sn ( t , α 2 ) = x 1 α x + α
as described in Byrd and Friedman.
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Kymani Hatfield
Answered 2022-10-18 Author has 2 answers
For any real number of x,
When | x | 1
J = x 1 2 ( 1 + x 2 ) 1 2
= n = 0 ( 1 ) n ( 2 n ) ! x 2 n + 1 2 4 n ( n ! ) 2 ( 1 2 n ) d x
= n = 0 ( 1 ) n ( 2 n ) ! x 2 n + 3 2 4 n ( n ! ) 2 ( 1 2 n ) ( 2 n + 3 2 ) + C
= n = 0 ( 1 ) n ( 2 n ) ! x 2 n + 3 2 4 n ( n ! ) 2 ( 1 2 n ) ( 2 n + 3 2 ) + C
= n = 0 ( 1 ) n ( 2 n ) ! x 2 n + 3 2 2 2 n 1 ( n ! ) 2 ( 1 2 n ) ( 4 n + 3 ) + C
I = ( 1 + x ) 1 2 ( 1 + x 2 ) 1 2
= ( x + 1 ) 1 2 n = 0 ( 1 ) n ( 2 n ) ! x 2 n 4 n ( n ! ) 2 ( 1 2 n ) d x
= ( x + 1 ) 1 2 n = 0 ( 1 ) n ( 2 n ) ! ( x + 1 1 ) 2 n 4 n ( n ! ) 2 ( 1 2 n ) d x
= n = 0 k = 0 2 n ( 1 ) n k ( ( 2 n ) ! ) 2 ( x + 1 ) k + 1 2 4 n ( n ! ) 2 k ! ( 2 n k ) ! ( 1 2 n ) d x
= n = 0 k = 0 2 n ( 1 ) n k ( ( 2 n ) ! ) 2 ( x + 1 ) k + 3 2 4 n ( n ! ) 2 k ! ( 2 n k ) ! ( 1 2 n ) ( k + 3 2 ) + C
= n = 0 k = 0 2 n ( 1 ) n k ( ( 2 n ) ! ) 2 ( x + 1 ) k + 3 2 2 2 n 1 ( n ! ) 2 k ! ( 2 n k ) ! ( 1 2 n ) ( 2 k + 3 ) + C
When | x | 1
J = x 1 2 ( 1 + x 2 ) 1 2
= x 1 2 ( x 2 ( 1 x 2 + 1 ) ) 1 2
= x 3 2 ( 1 + 1 x 2 ) 1 2
= x 3 2 n = 0 ( 1 ) n ( 2 n ) ! x 2 n 4 n ( n ! ) 2 ( 1 2 n ) d x
= n = 0 ( 1 ) n ( 2 n ) ! x 3 2 2 n 4 n ( n ! ) 2 ( 1 2 n ) d x
= n = 0 ( 1 ) n ( 2 n ) ! 2 2 n 1 ( n ! ) 2 ( 2 n 1 ) ( 4 n 5 ) x 2 n 5 2 + C
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