d x x^ 1 2 +x^ 1 3 = A x^ 1 2 + B x^ 1 3 + C x^ 1 6 + D (x^ 1 6 +1 )+ k (where k is the integration constant) then values of A, B, C and D are respectively,

Step-by-step solution

To evaluate dx x^ 1 2 + x^ 1 3 , substitute x = t^6 , which yields dx = 6t^5dt , x^ 1 2 = t^3 , and x^ 1 3 = t^2 . Substituting gives 6t^5dt t^3 + t^2 = 6t^3dt t+1 . Dividing t^3 by t+1 produces t^2 - t + 1 - 1 t+1 , so the integral becomes 6 (t^2 - t + 1 - 1 t+1 )dt . Integrating term by term yields 6( t^3 3 - t^2 2 + t - |t+1|) = 2t^3 - 3t^2 + 6t - 6 |t+1| . Substituting t = x^ 1 6 gives 2x^ 1 2 - 3x^ 1 3 + 6x^ 1 6 - 6 (x^ 1 6 +1) + k, using the fact that x^ 1 6 +1 > 0 for x > 0 . Matching to the form Ax^ 1 2 +Bx