d x (x+2) (x^2+1 ) =p |x+2|+q |x^2+1 |+r ⁻¹ x+c then p+q+r=

Step-by-step solution

Let 1 (x+2)(x^2+1) = A x+2 + Bx+C x^2+1 . Multiplying both sides by (x+2)(x^2+1) , we get 1 = A(x^2+1) + (Bx+C)(x+2) . Setting x = -2 , we get 1 = A(4+1) + 0 , which implies 5A = 1 , so A = 1 5 . Comparing coefficients of x^2 , we get A + B = 0 , so B = -A = - 1 5 . Comparing constant terms, we get A + 2C = 1 , so 1 5 + 2C = 1 , which implies 2C = 4 5 , so C = 2 5 . The integral becomes ( 1/5 x+2 + -x/5 + 2/5 x^2+1 ) dx = 1 5 dx x+2 - 1 10 2x x^2+1 dx + 2 5 dx x^2+1 . This evaluates to 1 5 |x+2| - 1 10 |x^2+1| + 2