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

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

[From: Mathematics] [author: ] [Date: 04-11] [Hit: ]
What is the general anti derivative of f(x)=6x^5+19x^3−(11/x)?......

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

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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
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Como say: :-
f ' (x) = 30 x⁴ + 57 x² + 22/x²
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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
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Jim Moor say: for ax^n ...
Differentiate as
anx^n-1 + ....

Integrate the opposite
ax^n + c
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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.