_ - ^ x x 1+ ^2 x d x=

Step-by-step solution

Let f(x)= x x 1+ ^2 x f(-x)= (-x) (-x) 1+ ^2 x = x x 1+ ^2 x f ( x )= f (- x ) f ( x ) is an even function Let I= _ - ^ x x 1+ ^2 x d x I =2 ₀^ x x 1+ ^2 x dx =2 ₀^ ( -x) ( -x) 1+[ ( -x)]^2 d x=2 ₀^ ( -x) x 1+ ^2 x d x Eq. (1) +(2) gives, 2 I=2 ₀^ x 1+ ^2 x Put x=t x d x=-d t . Also when x=0, t=1 and when aligned & x = , t =-1 & 2 I =2 ₁⁻¹ - dt 1+ t ^2 & =2 _ -1 ^1 dt 1+ t ^2 =4 ₀^1 dt 1+ t ^2 =4 [ ⁻¹ ]₀^2=4 ( 4 )= ^2 & I = ^2 2 aligned