integral x/(x-1) .dx = x - ln(x-1) + c
where ln = natural logarithm
and c = constant of integration
alternatively if you meant: integral x/x - 1 .dx = c
∫ f(x)/(1 - f(x)) dx = -x + ∫ 1/(1 - f(x)) dx
(x - 16)(x + 2) x = 16 or -2
In reimann stieltjes integral if we assume a(x) = x then it becomes reimann integral so we can say R-S integral is generalized form of reimann integral.
If you mean integral[(2x^2 +4x -3)(x+2)], then multiply them out to get: Integral[2x^3+8x^2+5x-6]. This is then easy to solve and is = 2/4x^4+8/3x^3+5/2x^2-6x +c
X= plus or minus 1
the integral of the square-root of (x-1)2 = x2/2 - x + C
- ln ((x^2)-4)
x-40
-22 - -22
irdk
Integral of [1/(sin x cos x) dx] (substitute sin2 x + cos2 x for 1)= Integral of [(sin2 x + cos2 x)/(sin x cos x) dx]= Integral of [sin2 x/(sin x cos x) dx] + Integral of [cos2 x/(sin x cos x) dx]= Integral of (sin x/cos x dx) + Integral of (cos x/sin x dx)= Integral of tan x dx + Integral of cot x dx= ln |sec x| + ln |sin x| + C
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.
∫ f(x)/(1 - f(x)) dx = -x + ∫ 1/(1 - f(x)) dx
integral (a^x) dx = (a^x) / ln(a)
By factoring I get x-3 divided by x+3
0
The expression "x minus one" represents the result of subtracting one from the variable x. It can also be written as x - 1. This expression is a binomial, a mathematical expression with two terms, in this case, x and -1. The value of "x minus one" will depend on the specific value of x given in the context of the problem or equation.