Evaluate the expression ∫23 dx1-x2

Given expression in the problem is,∫23 11-x2⋅dx=∫23 1(1-x)(1+x)dx…..(1) (since, a2-b2=(a+b)(a-b))Now apply partial fraction decomposition as, =11-x1+x=A1-x+B1+x1=A(1+x)+B(1-x) Put x=1 in above equation,1=A1+1+B1-1⇒1=2A ⇒A=12 Put x=-1 in above equation,1=A1-1+B1+1⇒1=2B ⇒B=12 Put the values of ‘A’ and ‘B’ in equation (2),11-x1+x=121-x+121+x ⇒11-x1+x=121(1-x)+1(1+x) Put this value in equation (i),∫23 11-x2dx=∫23 1211-x+11+xdx ⇒∫23 11-x2dx =12∫23 1(1-x)dx+∫23 1(1+x)dx Suppose, u=1-x⇒du=-dx, v=1+x⇒dv=dxThus,⇒∫23 11-x2dx=12∫23 1u(-du)+∫23 1vdv ⇒∫23 11-x2dx=12[-log⁡u+log⁡v]23 Put values of ‘u’ and ‘v’ again,⇒∫23 11-x2dx=12[-log⁡(1-x)+log⁡(1+x)]23 ⇒∫23 11-x2dx =12[log⁡(1+x)-log⁡(1-x)]23 ⇒∫23 11-x2dx=12log⁡(1+x)(1-x)23 (since, log⁡ab=log⁡a-log⁡b)Here we are applying the upper and lower bounds,⇒∫23 11-x2dx=12log⁡1+31-3-log⁡1+21-2⇒∫23 11-x2dx=12log⁡42-log⁡31⇒∫23 11-x2dx=12log⁡2-log⁡3⇒∫23 11-x2dx=12log⁡23 So, option 1 is correct.

Evaluate the expression ∫23 dx1-x2

Given expression in the problem is,∫23 11-x2⋅dx=∫23 1(1-x)(1+x)dx…..(1) (since, a2-b2=(a+b)(a-b))Now apply partial fraction decomposition as, =11-x1+x=A1-x+B1+x1=A(1+x)+B(1-x) Put x=1 in above equation,1=A1+1+B1-1⇒1=2A ⇒A=12 Put x=-1 in above equation,1=A1-1+B1+1⇒1=2B ⇒B=12 Put the values of ‘A’ and ‘B’ in equation (2),11-x1+x=121-x+121+x ⇒11-x1+x=121(1-x)+1(1+x) Put this value in equation (i),∫23 11-x2dx=∫23 1211-x+11+xdx ⇒∫23 11-x2dx =12∫23 1(1-x)dx+∫23 1(1+x)dx Suppose, u=1-x⇒du=-dx, v=1+x⇒dv=dxThus,⇒∫23 11-x2dx=12∫23 1u(-du)+∫23 1vdv ⇒∫23 11-x2dx=12[-log⁡u+log⁡v]23 Put values of ‘u’ and ‘v’ again,⇒∫23 11-x2dx=12[-log⁡(1-x)+log⁡(1+x)]23 ⇒∫23 11-x2dx =12[log⁡(1+x)-log⁡(1-x)]23 ⇒∫23 11-x2dx=12log⁡(1+x)(1-x)23 (since, log⁡ab=log⁡a-log⁡b)Here we are applying the upper and lower bounds,⇒∫23 11-x2dx=12log⁡1+31-3-log⁡1+21-2⇒∫23 11-x2dx=12log⁡42-log⁡31⇒∫23 11-x2dx=12log⁡2-log⁡3⇒∫23 11-x2dx=12log⁡23 So, option 1 is correct.