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I'm not sure if you mean e^x + 17 or e^(x+17) so we'll do both.

First, the integral of e^x + 17

because these terms are being added you can integrate them separately:

integral((e^x)dx) + integral(17dx)

integral of e^x is just e^x + C

Integral of 17 is 17x + C, so we get:

e^x + 17x + C

Second, the integral of e^(x+17)

we know how to integrate the form e^u, so just do a u substitution

u=x+17

du=dx

so we get

integral((e^u)du)=e^u + C

resubstitute for u and get e^(x+17) + C

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0One 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.

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

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

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

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

sqrt(1) + 3*sqrt(x) = 1 + 3*x^1/2So the antiderivative is x + [3*x^(3/2)]/(3/2) + c = x + 2*x^(3/2) + c where c is the constant of integration.

It is -exp (-x) + C.

If: x = -3x+1 Then: x+3x = 1 => 4x =1 So: x = 1/4 or 0.25 ----------- I notice that the question requests a solution for g x = -3x + 1. It seems possible that parentheses around the 'x' after the 'g' have gone missing, along with a prime indicating the derivative of the function g. This being the case, we would be seeking the antiderivative of -3x + 1. The antiderivative of a sum is the sum of the antiderivatives. So we can look at -3x and +1 separately. The derivative of x2 is 2x. Therefore, the antiderivative of x is x2/2, and the antiderivative of -3x is -3x2/2. The antiderivative of 1 is x. Overall, the solution is the antiderivative -3x2/2 + x + C, where C is an arbitrary constant.

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

x/(x+1) = 1 - 1/(x + 1), so the antiderivative (or indefinite integral) is x + ln |x + 1| + C,

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.

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.

X(logX-1) + C

4x + 17 = x + 2 4x - x = 2 - 17 3x = -15 x = -5

(2/3)*x^(3/2)

x + 17, perhaps.x + 17, perhaps.x + 17, perhaps.x + 17, perhaps.

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.

The anti-derivative of any constant c, is just c*x. Thus, the antiderivative of pi is pi*x. We can verify this by taking the derivative of pi*x, which gives us pi.

It is ln(ln(x))

sqrt(X + 1) is also; (X + 1)^1/2 ( add 1 to exponent and multiply by inverse ) 2/3(X +1)^3/2

The anti-derivative of sqrt(x) : sqrt(x)=x^(1/2) The anti-derivative is x^(1/2+1) /(1/2+1) = (2/3) x^(3/2) The anti-derivative is 4e^x is 4 e^x ( I hope you meant e to the power x) The anti-derivative of -sin(x) is cos(x) Adding, the anti-derivative is (2/3) x^(3/2) + 4 e^x + cos(x) + C

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