∫ xn dx = xn+1/(n+1) + C
(n ≠-1)
C is the constant of integration.
if you are integrating with respect to x, the indefinite integral of 1 is just x
∫ ax dx = ax/ln(a) + C C is the constant of integration.
d/dx ∫ f(x) dx = f(x)
With respect to x, this integral is (-15/2) cos2x + C.
What do you mean? As this is a calculus question, I presume that you are asking for a derivative or integral The derivative of any function of the form ƒ(x) = a * x ^ n is ƒ'(x) = a * n * x ^ (n-1) The integral of any function of the form ∫ a*x ^ n is a / (n+1) * x ^ (n+1) + C Your function that you gave is 1 / x^(2) which is equal to: x^(-2) Thus the derivative is: -2 * x^(-3) And the integral is: -x^(-1) + C
∫ f(x)nf'(x) dx = f(x)n + 1/(n + 1) + C n ≠-1 C is the constant of integration.
∫ f'(x)(af(x) + b)n dx = (af(x) + b)n + 1/[a(n + 1)] + C C is the constant of integration.
∫ ex dx = ex + CC is the constant of integration.
The integral of X 4Y X 8Y 2 With respect to X is 2ln(10/9).
if you are integrating with respect to x, the indefinite integral of 1 is just x
integral of e to the power -x is -e to the power -x
∫ f(x)/[f(x) + g(x)]n dx = ∫ 1/[f(x) + g(x)]n - 1 dx - ∫ g(x)/[f(x) + g(x)]n dx
The integral of the function 1 sinc(x) with respect to x is x - cos(x) C, where C is the constant of integration.
For n not equal to -1, it is 1/(n+1)*xn+1 while for n = -1, it is ln(|x|), the logarithm to base e.
∫ ax dx = ax/ln(a) + C C is the constant of integration.
d/dx ∫ f(x) dx = f(x)
With respect to x, this integral is (-15/2) cos2x + C.