The integral of 2-x = 2x - (1/2)x2 + C.
x^2/2
The integral of arcsin(x) dx is x arcsin(x) + (1-x2)1/2 + C.
I wasn't entirely sure what you meant, but if the problem was to find the integral of [sec(2x)-cos(x)+x^2]dx, then in order to get the answer you must follow a couple of steps:First you should separate the problem into three parts as you are allowed to with integration. So it becomes the integral of sec(2x) - the integral of cos(x) + the integral of x^2Then solve each part separatelyThe integral of sec(2x) is -(cos(2x)/2)The integral of cos(x) is sin(x)The integral of x^2 isLastly you must combine them together:-(cos(2x)/2) - sin(x) + (x^3)/3
x
The integral of 2-x = 2x - (1/2)x2 + C.
integral of (tanx)^4 (tanx)^4 = (tanx)^2 (tanx)^2 =(sec^2 x - 1)(tan^2 x) =(sec^2 x)(tan^2 x) - tan^2 x = integral of sec^2 x tan^2 x dx - integral of tan^2 x dx First, integral of sec^2 x tan^2 x dx Let u = tanx because that would make du = sec^2 x dx so then we have integral of u^2 du which is (1/3)u^3 substituting back in tanx we get (1/3)tan^3 x Next, integral of tan^2 x tan^2 x = sec^2 x -1 integral of sec^2 x - 1 = integral of sec^2 x dx - integral 1 dx = tanx - x so putting it all together we have integral of tan^4 x dx = (1/3)tan^3 x - tanx + x + C
By antiderivative do you mean integral? If yes, integral x^1 dx= (x^2)/2
Integral of x dx / sqrt(x+2) Make the substitution sqrt(x+2)=u (x+2)^(1/2) = u (1/2)(x+2)^(-1/2) dx = du 1/2(x+2)^(1/2) dx = du 1/2sqrt(x+2) dx = du 1/sqrt(x+2) dx = 2 du Integral of x dx / sqrt(x+2) = Integral 2 x du sqrt(x+2) = u (x+2)=u^2 x=u^2-2 Integral 2 x du = Integral 2(u^2-2) du = Integral 2u^2 du - 4 du = 2 u^3/3 - 4u + C = (2/3) (x+2)^(3/2) - 4 sqrt(x+2) + C
The integral of X 4Y X 8Y 2 With respect to X is 2ln(10/9).
x^2/2
In order to evaluate a definite integral first find the indefinite integral. Then subtract the integral evaluated at the bottom number (usually the left endpoint) from the integral evaluated at the top number (usually the right endpoint). For example, if I wanted the integral of x from 1 to 2 (written with 1 on the bottom and 2 on the top) I would first evaluate the integral: the integral of x is (x^2)/2 Then I would subtract the integral evaluated at 1 from the integral evaluated at 2: (2^2)/2-(1^2)/2 = 2-1/2 =3/2.
You add one to the power, and then divide by the power that it has so you would have: Integral of x = (x^2)/2 Integral of x^2 = (x^3)/3 Etc.
the integral of the square-root of (x-1)2 = x2/2 - x + C
The integral of arcsin(x) dx is x arcsin(x) + (1-x2)1/2 + C.
I wasn't entirely sure what you meant, but if the problem was to find the integral of [sec(2x)-cos(x)+x^2]dx, then in order to get the answer you must follow a couple of steps:First you should separate the problem into three parts as you are allowed to with integration. So it becomes the integral of sec(2x) - the integral of cos(x) + the integral of x^2Then solve each part separatelyThe integral of sec(2x) is -(cos(2x)/2)The integral of cos(x) is sin(x)The integral of x^2 isLastly you must combine them together:-(cos(2x)/2) - sin(x) + (x^3)/3
If x has the power 2 then you want the integral of x2, I think. When you integrate this you get : x3/3 , plus a constant.