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
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∫ f(x)/(1 - f(x)) dx = -x + ∫ 1/(1 - f(x)) dx
(x - 16)(x + 2) x = 16 or -2
X= plus or minus 1
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