Indefinite integrals question?

Question

Find the following indefinite integral:

∫x/1-x^2 dx

Please show your work and explain how it is done.

Thanks,

Answer 1

∫x/(1-x^2) dx

let u = 1-x^2
du= -2xdx

∫x/(1-x^2) dx
=(-1/2)∫(-2x)/(1-x^2) dx
=(-1/2)∫1/u du
= -(1/2)ln(u) + c
= -(1/2)ln(1-x^2) + c

Answer 2

∫x/1-x^2 dx
I would use the substitution method:
u = 1-x^2 therefore 1/1-x^2 = 1/u = u^-1
du = -2x *dx
(-1/2) du = x * dx
Then we get:
∫x/1-x^2 dx
= ∫(-1/2) (u^-1) du
= (-1/2) ∫(u^-1) du
= (-1/2) * ln u + c // properties of log: a*lnb = ln(b^a)
= ln (u^-1/2) + c
= ln [1/sqrt(u)] + c
= ln [1/sqrt(1-x^2)] + c

Answer 3

Let u = (1- x^2) then du = -2xdx That should be enough to get you on your way