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What is the integral of the derivative with respect to x of the function f divided by the square root of the quantity a times f plus b with respect to x?
Mrkbh
∫ f'(x)/√(af(x) + b) dx = 2√(af(x) + b)/a + C
C is the constant of integration.
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∙ 13y ago
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What is the integral of the derivative with respect to x of a function of x divided by that same function of x with respect to x?
∫ f'(x)/f(x) dx = ln(f(x)) + C C is the constant of integration.
What is the integral of the derivative with respect to x of the function f divided by the square root of the quantity f squared plus a constant with respect to x?
∫ f'(x)/√[f(x)2 + a] dx = ln[f(x) + √(f(x)2 + a)] + C C is the constant of integration.
What is the integral of the derivative with respect to x of f divided by the quantity p squared plus q squared f squared with respect to x where f is a function of x and p and q are constants?
∫ f'(x)/(p2 + q2f(x)2) dx = [1/(pq)]arctan(qf(x)/p)
What is the integral of the derivative with respect to x of f divided by the quantity q squared f squared minus p squared with respect to x where f is a function of x and p and q are constants?
∫ f'(x)/( q2f(x)2 - p2) dx = [1/(2pq)ln[(qf(x) - p)/(qf(x) + p)]
What is the integral of f divided by the quantity 1 minus f with respect to x where f is a function of x?
∫ f(x)/(1 - f(x)) dx = -x + ∫ 1/(1 - f(x)) dx
Equation for marginal cost and average cost?
Marginal cost - the derivative of the cost function with respect to quantity. Average cost - the cost function divided by quantity (q).
What is the integral of the derivative with respect to x of a function of x divided by that same function of x with respect to x?
∫ f'(x)/f(x) dx = ln(f(x)) + C C is the constant of integration.
What is the integral of the derivative with respect to x of the function f divided by the square root of the quantity f squared plus a constant with respect to x?
∫ f'(x)/√[f(x)2 + a] dx = ln[f(x) + √(f(x)2 + a)] + C C is the constant of integration.
What is the integral of the derivative with respect to x of f divided by the quantity p squared plus q squared f squared with respect to x where f is a function of x and p and q are constants?
∫ f'(x)/(p2 + q2f(x)2) dx = [1/(pq)]arctan(qf(x)/p)
What is the integral of the quantity of the derivative with respect to x of the function f times another function of x defined as g subtracted by g prime times f divided by g squared with respect to x?
∫ [f'(x)g(x) - g'(x)f(x)]/g(x)2 dx = f(x)/g(x) + C C is the constant of integration.
What is the integral of the derivative with respect to x of f divided by the quantity q squared f squared minus p squared with respect to x where f is a function of x and p and q are constants?
∫ f'(x)/( q2f(x)2 - p2) dx = [1/(2pq)ln[(qf(x) - p)/(qf(x) + p)]
What is the integral of the quantity of the derivative with respect to x of the function f times another function of x defined as g subtracted by g prime times f divided by f times g with respect to x?
∫ [f'(x)g(x) - g'(x)f(x)]/[f(x)g(x)] dx = ln(f(x)/g(x)) + C C is the constant of integration.
What is the integral of f divided by the quantity 1 minus f with respect to x where f is a function of x?
∫ f(x)/(1 - f(x)) dx = -x + ∫ 1/(1 - f(x)) dx
How do you integrate the integral of quantity of x plus 2 divided by x squared plus x plus 1?
3
What is the integral of 1 divided by the quantity of a function of x multiplied by the quantity of that same function plus or minus another function of x with respect to x?
∫ [1/[f(x)(f(x) ± g(x))]] dx = ±∫1/[f(x)g(x)] dx ± (-1)∫ [1/[g(x)(f(x) ± g(x))]] dx
What is the integral of f prime divided by the quantity f times the square root of the quantity f squared minus a squared with respect to x where f is a function of x and a is a constant?
∫ f'(x)/[f(x)√(f(x)2 - a2)] dx = (1/a)arcses(f(x)/a) + C C is the constant of integration.
What is the derivative of ln1 divided by x?
Given y=ln(1/x) y'=(1/(1/x))(-x-2)=(1/(1/x))(1/x2)=x/x2=1/x Use the chain rule. The derivative of ln(x) is 1/x. Instead of just "x" inside the natural log function, it's "1/x". Since the inside of the function is not x, the derivative must be multiplied by the derivative of the inside of the function. So it's 1/(1/x) [the derivative of the outside function, natural log] times -x-2=1/x2 [the derivative of the inside of the function, 1/x] This all simplifies to 1/x So the derivative of ln(1/x) is 1/x
