integral:s(ax^2+bx^3)/(root of x)*dx?

Answer 1

Representing integral as S:

S[(ax^2+bx^3) / (root of x)]*dx

Put x = t²

dx = 2t*dt

Thus : S[(at^4+bt^6) / t]*2t*dt

= S(2(at^4+bt^6))dt

= 2a*S(t^4) + 2b*S(t^6)

= 2a*t^5/5 + 2b*t^7/7 + c

= (2a/5)t^5 + (2b/7)t^7 + c

where c is the constant of integration

put t =root x and get the answer in terms of 'x'

= (2a/5)x^(5/2) + (2b/7)t^(7/2) + c'

where c' is the constant of integration

Answer 2

(ax^2 + bx^3)/(root x) can be written as
ax^2/x^(1/2) + bx^3/x^(1/2)
= ax^(3/2) + bx^(5/2)
Now integrate with respect to x (remember that you add one to the exponent and then divide by that new exponent).
integral of [ax^(3/2) + bx^(5/2)]dx
= ax^(5/2) / (5/2) + bx^(7/2) / (7/2) + C
= (2a/5)x^(5/2) + (2b/7)x^(7/2) + C

Answer 3

(5*a/2)*x^(5/2)+(7*b/2)x^(7/2)