Q: What is the antiderivative of e to the -x?

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One can use integration by parts to solve this. The answer is (x-1)e^x.

The integral would be 10e(1/10)x+c

You can't, unless it's an initial value problem. If f(x) is an antiderivative to g(x), then so is f(x) + c, for any c at all.

If f' (x) = x43, then f(x) = (1/44) x44 + C.

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One can use integration by parts to solve this. The answer is (x-1)e^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.

The antiderivative, or indefinite integral, of ex, is ex + C.

The integral would be 10e(1/10)x+c

Antiderivative of x/-1 = -1(x^2)/2 + C = (-1/2)(x^2) + C Wolfram says antiderivative of x^-1 is log(x) + C

1/ln(x)*e^(1/x) if you differentiate e^(1/x), you will get ln(x)*e^(1/x). times this by 1/ln(x) and you get you original equation. Peace

By antiderivative do you mean integral? If yes, integral x^1 dx= (x^2)/2

(that weird integral or antiderivative sign) x^(-6/5) dx =-5*x^(-1/5)

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.

It is -exp (-x) + C.

The general formula for powers doesn't work in this case, because there will be a zero in the denominator. The antiderivative of 1/x is ln(x), that is, the natural logarithm of x.

int(e 3x) = (1/3)e 3x ========