0
∫ 4/x dx
= 4 ∫ 1/x dx
= 4ln(x) + C
This is true for three reasons:
- the derivative of the term ln(x) is equal to 1/x
- 4 is a constant factor of the term, and can be moved out of the integral
- C is an unknown constant, because we're looking at an indefinite integral
You can confirm this by taking the derivative of 4ln(x), which gives you 4/x, the original term.
What else can I help you with?
What is the integral of x divided by x plus 1?
x/(x+1) = 1 - 1/(x + 1), so the antiderivative (or indefinite integral) is x + ln |x + 1| + C,
What is the antiderivative of logx?
X(logX-1) + C
What is the antiderivative of square root of x?
(2/3)*x^(3/2)
X-2 times x plus 4 divided by x squared - 4?
x-2(x)+4/x^2 -4=x-2x+4/x^2 -4=-x-4+4/x^2
Y equals 4 divided by x squared what formula is this?
Y = 4/x^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 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 1 divided by 18x5?
-1/72x4 + c where c is the constant of integration.
What is the integral of x divided by x plus 1?
x/(x+1) = 1 - 1/(x + 1), so the antiderivative (or indefinite integral) is x + ln |x + 1| + C,
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.
How do you solve g x equals -3x plus 1?
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.
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)
Can anyone of can prove that antiderivative is wrong?
The way to disprove an antiderivative is to simply differentiate the function and see if it matches the integral expression. Remember that an antiderivative expression must include a term often coined "C-" an arbitrary constant. For example, ∫(x^3 +14x)dx= (1/4)X^4+ 7X^2 +C. To verify that this is correct, take the derivative. You get x^3 +14x.
What is the antiderivative of exp -x?
It is -exp (-x) + C.
What is the antiderivative of e to the -x?
-e-x + C.
How do you take the antiderivative of 1 over x?
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.
