2 x (x^2+1 ) (x^2+2 )^2 d x=

Step-by-step solution

Let I = 2x (x^2+1)(x^2+2)^2 dx . Substitute t = x^2 , then dt = 2x dx . The integral becomes I = dt (t+1)(t+2)^2 . Using partial fractions, let 1 (t+1)(t+2)^2 = A t+1 + B t+2 + C (t+2)^2 . Multiplying by (t+1)(t+2)^2 , we get 1 = A(t+2)^2 + B(t+1)(t+2) + C(t+1) . For t = -1 , 1 = A(1)^2 A = 1 . For t = -2 , 1 = C(-1) C = -1 . Comparing coefficients of t^2 , 0 = A + B B = -A = -1 . Thus, I = ( 1 t+1 - 1 t+2 - 1 (t+2)^2 ) dt . Integrating, I = |t+1| - |t+2| - ( - 1 t+2 ) + c = | t+1 t+2 | + 1 t+2 + c . Substituting t