0
ex and ln(x) are inverse functions.
With this you can get 5x = eln(5^x)
Therefore you can anti-differentiate this to get eln(5^x)/(ln(5x))
Which equals 5x/ln(5x)
Wiki User
∙ 13y ago
Add your answer:
Earn +20 pts
What is the antiderivative of x to the negative 6 5ths?
(that weird integral or antiderivative sign) x^(-6/5) dx =-5*x^(-1/5)
What is the antiderivative of x raised to the pi power?
x(pi+1)/(pi+1)
What is the antiderivative of x to the 1?
By antiderivative do you mean integral? If yes, integral x^1 dx= (x^2)/2
What is the antiderivative of e to the power of one divided by -x?
Powers of e are simple to integrate. The derivative of eu equals u'eu; inversely, the antiderivative of eu equals eu/u'. Therefore, the antiderivative of e1/-x equals (e1/-x)/{d/dx[1/-x]}. The derivative of 1/-x, which can also be expressed as x-1, equals (-1)x(-1-1) = -x-2 = -1/x2.
What is the antiderivative of -10x4?
I assume you mean -10x^4? In that case, antiderivative would be to add one to the exponent, then divide by the exponent. So -10x^5, then divide by 5. So the antiderivative is -2x^5.
What is the antiderivative of x to the negative 6 5ths?
(that weird integral or antiderivative sign) x^(-6/5) dx =-5*x^(-1/5)
What is the antiderivative of x raised to the pi power?
x(pi+1)/(pi+1)
Integral of e to the power of x?
The antiderivative, or indefinite integral, of ex, is ex + C.
What is the antiderivative 7x5-Cos x?
If the first term is 7x^5, ∫7x^5 -cox dx is the expression. You can split this up into two integrals if that helps you visualize the terms. ∫7x^5dx - ∫cox dx. We know that the antiderivative of cosx is sinx, so that is our second term. In the first term, we must undo the power rule, adding one to the power and multiplying by the reciprocal of the power. This gives us (7/6)x^6. So, our final antiderivative expression is (7/6)x^6-sinx+C, with C being an arbitrary constant.
What is the antiderivative of x to the -1?
Antiderivative of x/-1 = -1(x^2)/2 + C = (-1/2)(x^2) + C Wolfram says antiderivative of x^-1 is log(x) + C
What is the antiderivative of x to the 1?
By antiderivative do you mean integral? If yes, integral x^1 dx= (x^2)/2
What is the antiderivative of e to the power of one divided by -x?
Powers of e are simple to integrate. The derivative of eu equals u'eu; inversely, the antiderivative of eu equals eu/u'. Therefore, the antiderivative of e1/-x equals (e1/-x)/{d/dx[1/-x]}. The derivative of 1/-x, which can also be expressed as x-1, equals (-1)x(-1-1) = -x-2 = -1/x2.
What is the antiderivative of e to the -x?
-e-x + C.
What is the antiderivative of exp -x?
It is -exp (-x) + C.
What is the antiderivative of -10x4?
I assume you mean -10x^4? In that case, antiderivative would be to add one to the exponent, then divide by the exponent. So -10x^5, then divide by 5. So the antiderivative is -2x^5.
What is the antiderivative of -1x 1?
If the term is -x, the integral expression is simply -∫x. By undoing the power rule, we get -(1/2)x^2+C, an arbitrary constant.
Is there any other anti derivative of 1 divided by x?
The antiderivative of 1/x is ln(x) + C. That is, to the natural (base-e) logarithm, you can add any constant, and still have an antiderivative. For example, ln(x) + 5. These are the only antiderivatives; there are no different functions that have the same derivatives. This is valid, in general, for all antiderivatives: if you have one antiderivative of a function, all other antiderivatives are obtained by adding a constant.
