What is the integral of x over 1 minus x squared?

Answer 1

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.