1.001
x.001 / .001
= 1000 x.001
I will assume that this is sopposed to be integrated with respect to x. To make this problem easier, imagine that the integrand is x raised to the negative 3. The integral is 1/(-2x-2) plus some constant c.
∫ 1/cos(x) dx = ln(sec(x) + tan(x)) + C C is the constant of integration.
The integral of -x2 is -1/3 x3 .
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.
The integral of arcsin(x) dx is x arcsin(x) + (1-x2)1/2 + C.
-(x-1)-1 or -1/(x-1)
.0999 rounded to the nearest tenth is .1
1 to any integral power is 1.
The indefinite integral of (1/x^2)*dx is -1/x+C.
3
0.5
0.5
x/(x+1) = 1 - 1/(x + 1), so the antiderivative (or indefinite integral) is x + ln |x + 1| + C,
3
If it is to a (-) power then that is the same as saying it is 1 divided by 2 to the power of 1. 2 to the power of 1 is 2, and so 1 divided by 2 is 1/2 or 0.5.
∫ (1/x) dx = ln(x) + C C is the constant of integration.
for expanding negative powers of a number you take 1 divided by the number to the positive power and expand. For example 2 to the -1 power is 1 divided by 2 to the + 1 power = 1/2 2 to the -3 power is 1 divided by 2 tot he + 3 power = 1/8 this is called inverse