_ n 1 n [ 1 n ⁻¹ 1 n + 2 n ⁻¹ 2 n + + 2 ]=

Step-by-step solution

Given that, _ n 1 n [ 1 n ⁻¹ 1 n + 2 n ⁻¹ 2 n + + 2 ] The above expression may be written as = _ n _ r=1 ^n r n ⁻¹ r n n = ₀^1 II x _ I ⁻¹ x d x Solving integration by parts method, we get = ⁻¹ x ₀^1 x d x- ₀^1 ( d d x ( ⁻¹ x ) x d x ) d x= [ ⁻¹ x x^2 2 ]₀^1- ₀^1 1 1-x^2 x^2 2 d x= [ ⁻¹ x x^2 2 ]₀^1- 1 2 ₀^1 x^2-1+1 1-x^2 d x= [ ⁻¹(1) 1 2 - ⁻¹(0) 0^2 2 ]- 1 2 ₀^1 - (1-x^2 ) 1-x^2 d x- 1 2 ₀^1 d x 1-x^2 = [ 2 1 2 -0 ]+ 1 2 ₀^1 1-x^2 d x- 1 2 ₀^1 d x 1-x^2 = 4 + 1 2 [ x 2 1-x^2 + 1 2 ⁻¹ x ]₀^1- 1 2 [ ⁻¹ x ]₀^1 [ a^2-