( ₀^ 1 2 d x (1+X² ) 1-X² ) is equal to

Step-by-step solution

Given that, ( ₀^ 1 / 2 d x (1+x² ) 1- ² ) Let ( x= ) then ( d x= d ) and limits become ( 0 0, 1 2 6 . ) So, [ array l ₀^ / 6 d (1+ ² ) 1- ² = ₀^ / 6 d (1+ ² ) = ₀^ / 6 ² 1+2 ² d array ] Let ( u= 2 ) then ( d u= 2 ² d ) and limits becomes ( 0 0, 6 2 3 ). So, ( 1 2 ₀^ 2 / 3 1 1+u² d u ) We know that ( 1 1+x² d x= ⁻¹ . ) So, [ array l . 1 2 ⁻¹ u |₀ ^ 2 / 3 = 1 2 ⁻¹ ( 2 3 )-0 = 1 2 ⁻¹ ( 2 3 ) array ]