.2x^5+x+C
f dx ?? do you mean f df ? int(f df) (1/2)f2 + C --------------
The integral of cot(x)dx is ln|sin(x)| + C
dy/dx = 3 integral = (3x^2)/2
∫e^(-2x) dx Let u = 2x du= 2 dx dx=(1/2) du ∫e^(-2x) dx = (1/2) ∫e^-u du = (1/2) (-e^-u) = -e^-u /2 + C = -e^-(2x) / 2 + C
integrate(x5x dx) simplifies to integrate(5x^2 dx), and using the power rule of integration, add one to the power of x and divide the term by that number. Thus, x5x dx integrated is (5/3)x^3
.2x^5+x+C
∫(4x3 - 2x2 + x - 1) dx You can integrate this by taking the antiderivative of each term. Each of these terms is in the format axn, the antiderivative of which is axn-1/n: = ∫(4x3)dx - ∫(2x2)dx + ∫(x)dx - ∫(1)dx = x4 - 2x3/3 + x2/2 - x + C
f dx ?? do you mean f df ? int(f df) (1/2)f2 + C --------------
Simply integrate all the pieces apart, en add them up. This is allowed, because int_a^c f(x)dx = int_a^b f(x)dx + int_b^c f(x)dx for all a,b,c in dom(f).
The integral of cot(x)dx is ln|sin(x)| + C
Integrate(0->t) (2-2x) dx is the integral correct (Integrate(0->t) (2-2x) dy would be different, you must state integrate with respect to what otherwise it can be anything) So integrals preserves sum and product with constants. i.e. Integrate (2-2x) dx = Integrate 2 dx - 2integrate x dx = 2x - x^2 + C By Fundamental Theorem of Calculus, take any anti-derivative, say C = 0 would be fine, and Integral(0->t)(2-2x) dx = (2x-x^2)|(0->t) = (2t-t^2) - 0 = 2t-t^2 It is a special case of the Second Fundamental Theorem of Calculus -- integral(0->t) f(x) dx is an anti-derivative of f(x).
so the problem is ∫(sin7x)dx you have to use u-substitution, set u=7x then du/dx=7 so du=7dx and solve for dx 1/7du=dx, so you get 1/7∫(sinu)du then integrate like normal, and you get 1/7(-cosu)+C then you plug u back in and get 1/7(-cos7x)+C
∫[√(4x) / x] dx = ∫(2 / √x)dx = 2∫(x-1/2) dx = 2(2x1/2 + C) = 4√x + C
dy/dx = 3 integral = (3x^2)/2
The integral for csc(u)dx is -ln|csc(u) + cot(u)| + C.
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