What is the general anti derivative of f(x)=6x^5+19x^3−(11/x)?

answers:
la console say: f(x) = 6x^(5) + 19x^(3)  (11/x)
f(x) = 6.[x^(5)] + 19.[x^(3)]  11.[1/x]
The derivative of u^(n) is: n * u' * n^(n  1)
The derivative of Ln(x) = 1/x
F(x) = 6.(1/6).[x^(6)] + 19.(1/4).[x^(4)]  11.Ln(x) + k
F(x) = x^(6) + (19/4).x^(4)  11.Ln(x) + k → where k is a constant

Como say: :
f ' (x) = 30 x⁴ + 57 x² + 22/x²

llaffer say: f(x) = 6x⁵ + 19x³  11/x
Let's put the x into negative exponent form, for now:
f(x) = 6x⁵ + 19x³  11x⁻¹
If you are getting the derivative of a polynomial, such as:
f(x) = axⁿ
For every term, the new coefficient is the product of the old coefficient and the old exponent, then you subtract 1 from the exponent:
f'(x) = (an)xⁿ⁻¹
To get the antiderivative, you reverse the process: Add one to the exponent, then divide the old coefficient by the new exponent to get the new coefficient:
f'(x) = axⁿ
f(x) = a/(n + 1) * xⁿ⁺¹
But the x⁻¹ terms becomes x⁰ and then you can't divide by 0, so this ends up being the coefficient times the natural log of x.
So do this for every term, then add a constant term that is unknown:
f(x) = 6x⁵ + 19x³  11x⁻¹
∫ f(x) = x⁶ + (19/4)x⁴  11 ln(x) + c

Jim Moor say: for ax^n ...
Differentiate as
anx^n1 + ....
Integrate the opposite
ax^n + c

rotchm say: Integrate term by term. And you know the anti derivative of xⁿ ...
But whats weird is that in the past you were asking about solving DE's.
Now you are regressing and asking about *simple* integrals...
Hopefully no one will spoil you the answer thereby depriving you from your personal enhancement; that would be very inconsiderate of them.

JD say: x^6 + (19/4) x^4  11lnx + C
