If 2 x+3 (x-1) (x^2+1 ) d x= _e (x-1)^ 5 2 (x^2+1 )^a - 1 2 ⁻¹ x+ A where A is an arbitrary constant, then the value of a is

Step-by-step solution

To evaluate 2x+3 (x-1)(x^2+1) dx , begin with partial fraction decomposition: 2x+3 (x-1)(x^2+1) = B x-1 + Cx+D x^2+1 Equating numerators gives: 2x+3 = B(x^2+1) + (Cx+D)(x-1) 2x+3 = (B+C)x^2 + (-C+D)x + (B-D) Comparing coefficients: B+C = 0 , -C+D = 2 , B-D = 3 Solving yields B = 5 2 , C = - 5 2 , D = - 1 2 The decomposition becomes: 5 2(x-1) - 5x+1 2(x^2+1) The integral evaluates to: 5 2 |x-1| - 1 2 5x+1 x^2+1 dx = 5 2 |x-1| - 1 2 (5 x x^2+1 dx + 1 x^2+1 dx ) = 5 2 |x-1| - 5 4 (x^2+1) - 1 2 ⁻¹x + A Rewriting using