X(logX-1) + C
(2/3)*x^(3/2)
If in the real number universe, first k is to be >0, y=kx = exlog(k) the antiderivative of eax is eax/a so the antiderivative of Y is exlog(k) / log(k) = kx /log(k)
x/(x+1) = 1 - 1/(x + 1), so the antiderivative (or indefinite integral) is x + ln |x + 1| + C,
∫ 4/x dx= 4 ∫ 1/x dx= 4ln(x) + CThis is true for three reasons:the derivative of the term ln(x) is equal to 1/x4 is a constant factor of the term, and can be moved out of the integralC is an unknown constant, because we're looking at an indefinite integralYou can confirm this by taking the derivative of 4ln(x), which gives you 4/x, the original term.
The antiderivative of 2x is x2.
1/2 of sin(2x)
(cosx)^2-(sinx)^2
The antiderivative of a function which is equal to 0 everywhere is a function equal to 0 everywhere.
35x2
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.
Antiderivative of x/-1 = -1(x^2)/2 + C = (-1/2)(x^2) + C Wolfram says antiderivative of x^-1 is log(x) + C
By antiderivative do you mean integral? If yes, integral x^1 dx= (x^2)/2
(x+sinxcosx)/2, can do it by parts or by knowing your double angle formulas
x can go to + or - infinity. f(x) is limited from + 1/2 to - 1/2.
The fundamental theorum of calculus states that a definite integral from a to b is equivalent to the antiderivative's expression of b minus the antiderivative expression of a.
(that weird integral or antiderivative sign) x^(-6/5) dx =-5*x^(-1/5)