Assuming it is int((x/(1-x2)dx)) then the answer is -.5ln|1-x2|+C. You have to use "u" substitution and isolate the 1-x2 and make it "u". A negative one-half is put out in front when you derive the value of "u" and the equation is -.5int(du/u). The integral of one over "u" is ln|u|+C. Substitute the "u" value back in and the answer becomes -.5ln|1-x2|+C.
I apologize for the impolication that I'm "improving" on Rusty's answer, not my intent.
Instead I'd like to point out that when you write "x over 1 minus x squared," I don't know if you mean:
x/1 - x^2 (it's a safe bet this isn't it, but your notation isn't clear)
x/(1-x)^2
x/(1-x^2) This is the one Rusty explained.
Writing your question as I've done above, or even using parentheses in plain English, e.g., "1 - (1 minus x) squared" will clarify your question.
Use u substitution:
u = 1-x -> x = 1+u
du = -dx -> dx = -du
x/(1-x)^2 dx = (1+u) / (u^2) = [1/u^2 + 1/u] * (-du)
= -du/u^2 + du/u
Int[-du/u^2 du] = -(-1/u) = 1/u
Int[du/u] = ln(u)
Now change back to x:
1/u + ln(u) = 1/(1-x) + ln(1-x) + C <- this is your answer.
Is this right? Let's take the derivative and find out.
d/dx (1/(1-x)) = d/dx ((1-x)^(-1)) = -1*(1-x)^(-2)) * -1 = 1/(1-x)^2
d/dx (ln(1-x)) = -1/(1-x) = -(1-x) / (1-x)^2
Adding these gives:
(1 - (1-x)) / (1-x)^2 = x / (1-x)^2
So the answer's correct.
No, unless "a" happens to be equal to 0, or to 1.
1 over x2 - 4 is the multiplicative inverse of x2 minus four 1/x2 - 4
(2x - 1)(4x + 1)
It is (A-1)(A+1) when factored
0.5
the integral of the square-root of (x-1)2 = x2/2 - x + C
arctan(x)
pi squared
5x lolzz 8)
No, unless "a" happens to be equal to 0, or to 1.
sin squared
-1² = 1
1 over x2 - 4 is the multiplicative inverse of x2 minus four 1/x2 - 4
∫1/(1-x)2dxWe can rewrite the integral as ∫(1-x)-2dx.Thus:∫(1-x)-2dxu = 1-xdu = -1-1∫-1(1-x)-2dx-1∫u-2du(u-1|0u=1-x(1-x)-1So we conclude that ∫1/(1-x)2dx = (1-x)-1 or 1/(1-x)
Multiply both sides by sin(1-cos) and you lose the denominators and get (sin squared) minus 1+cos times 1-cos. Then multiply out (i.e. expand) 1+cos times 1-cos, which will of course give the difference of two squares: 1 - (cos squared). (because the cross terms cancel out.) (This is diff of 2 squares because 1 is the square of 1.) And so you get (sin squared) - (1 - (cos squared)) = (sin squared) + (cos squared) - 1. Then from basic trig we know that (sin squared) + (cos squared) = 1, so this is 0.
The indefinite integral of (1/x^2)*dx is -1/x+C.
2x squared minus 5x minus 3 factored is (2x+1)(x-3).