Find the value of ₀¹ (1+x) 1+x² d x is

Step-by-step solution

Let I= ₀¹ (1+x) 1+x² d x Put x= d x= ² d When x=0, =0 and when x=1, = / 4 aligned ₀^ / 4 & (1+ ) 1+ ² ( ² ) d I &= ₀^ / 4 (1+ ) d ...(i) =& ₀^ / 4 [1+ ( 4 - ) ] d =& ₀^ / 4 [1+ / 4- 1+ / 4 ] d &= ₀^ / 4 (1+ 1- 1+ ) d =& ₀^ / 4 ( 2 1+ ) d aligned ₀^ / 4 [ 2- (1+ )] d ...(ii) On adding Eqs. (i) and (ii), we get aligned 2 I &= ₀^ / 4 2 d &= 2 ₀^ / 4 ld = ( )₀^ / 4 &= ( / 4-0)= 4 2 I &= 8 2 aligned