The value of 8 2 ₀^1 ( 1- x ^2 1+ x ^2 ) dx 1+ x ^4 is __________ .

Step-by-step solution

Let = 1+ x ^4 1+ x ^2 , we get aligned & - d d x = 2 x^3 (1+x^2 ) (1+x^4 ) - 2 x 1+x^4 (1+x^2 )^2 = 2 x^3 (1+x^2 )-2 x (1+x^4 ) (1+x^2 )^2 1+x^4 & = -2 x (1-x^2 ) (1+x^2 )^2 1+x^4 = 1-x^2 (1+x^2 ) 1+x^4 d x= ( 1+x^2 2 x ) d aligned The factor on the right can be put in terms of as follows aligned & 1+x^2 2 x = 1+x^4+2 x^2 4 x^2 = ( 1+x^4 2 x )^2+ 1 2 = ( ( 1+x^2 2 x ) )^2+ 1 2 & (1- ^2 ) ( 1+x^2 2 x )^2= 1 2 1+x^2 2 x = 1 2 aligned the given integral becomes 8 2 ₀^ / 4 d 2 = 8 2 1 2 ₀^ / 4 d = 8 2 1 2 4 =2