1 4 x 2 x d x= 1 2 2 ( 1+f(x) 1-f(x) ) - 1 2 g(x)+C , then g ( 6 )- 2 f ( 6 )=

Step-by-step solution

Let aligned I & = 1 4 x 2 x d x & =2 1 2 4 x 2 x d x & =2 1 6 x+ 2 x d x & =2 1 4 ^3 2 x-2 2 x d x aligned aligned & =2 1 2 2 x (2 ^2 2 x-1 ) d x & = 1 2 x (2 ^2 2 x-1 ) & Let 1 2 x (2 ^2 2 x-1 ) = A 2 x + B( 2 x)+C 2 ^2 2 x-1 & 1=(2 A+B) ^2 2 x+C 2 x-A & A=-1, B=2 and C=0 & I= 1 2 x (2 ^2 2 x-1 ) d x & = ( -1 2 x + 2 2 x 2 ^2 2 x-1 ) d x & I= - 2 x d x+ 2 2 x 1-2 ^2 2 x d x & I= -1 2 | 2 x+ 2 x|+I₁ & aligned Now, I₁= 2 2 x d x 1-2 ^2 2 x Let 2 x=t aligned & 2 2 x d x=d t & I₁= d t 1-2 t^2 = 1 2 d t ( 1 2 )^2-t^2 &