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integration of 1/1+x^4dx from 0 to infinity
Asked by poojabishnoipooja13 | 03 Dec, 2023, 06:59: PM
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(x4 + 1 ) = ( x2 + √2 x +1 ) ( x2 - √2 x +1 )
Hence we get , (Ax+B) ( x2 -√2 x + 1 ) + ( Cx+D) ( x2 + √2 x + 1) = 1
(A+C)x3 + ( B -√2 A + D + √2C ) x2 + ( A - √2B + C +√2D ) x + (B+D) = 1
By comparing constant term , B+D = 1
By comparing coefficients of x3 , we get , A+C = 0
By comparing coefficients of x2 and using B+D = 1, we get A-C = 1/√2
By comparing coefficients of x and using A+C = 0 , we get B-D = 0
Hence we get, A = 1/ (2√2 ) , B = 1/2 , C = - 1/(2√2) and D = 1/2
By using the values of A, B, C and D , we rewrite the following expression
Hence, we get
Above integration is split into two integrations I1 and I2 .
First integration is evaluated as
Second integration is evaluated as
After combining both integration , we get
Logarithmic term with lower limit becomes log (1) , hence vanishes .
logarithmic term with upper limit is calculated as follows
Hence we get
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