Integration again... ∫ x / [1- x² + √(1-x²) ] dx?

Question

So basically, I am stuck on another problem..if my notation is unclear, I'll try to explain it w/ words.
integral of x divided by 1 - x^2 + sqrt of (1-x^2)

thanks z o r r o!! i can't believe i overlooked that..the whole denominator thing threw me off. thanks a lot!

Answer 1

Let u = √(1-x²)

Then after taking the derivative of u and solving for x dx
we get

x dx = -u du

After the substitution you have:

Integral of - u du / (u^2 + u) = Integral of -du / (u+1)

= - Ln (u+1) +C

Susbtitute back.

-Ln (1+ √(1-x²))+C

Answer 2

∫ x / [1- x² + √(1-x²) ] dx =
- (√1 - x ²)(( √1 - x ²)+1) log (( √1 - x²)+1) / ( -x ² + (√1 - x ²) +1 + c

Good Luck Dear.

Answer 3

let u = 1- x ^2

du = -2xdx

∫ x / [1- x² + √(1-x²) ] dx
= -1/2∫ 1/(u+rt(u))du
= -1/2∫ 1/(rt(u)(rt(u)+1))du

another sub, let s = rtu, so ds = 1/(2rtu)du
-1/2∫ 1/(rt(u)(rt(u)+1))du
= -∫ 1/(s+1)ds
=-ln(s+1)+C

=-ln(rt(u)+1)+C
= -ln(rt(1-x^2)+1)+C

not too bad, just involved multiple subs.

Answer 4

u=sqrt(1-x^2) substitution =>

du/dx=-2*x/(2*sqrt(1-x^2^))=>

dx=-du*2*sqrt(1-x^2^)/2*x=>

dx=-sqrt(1-x^2^)/x*du

intgrl[x/(1-x^2+sqrt(1-x^2^)]dx=

-intgrl[[x/(u^2+u)*u*x*du=

-intgrl[u/(u+u^2)du=

-intgrl[1/(1+u)du=

-Ln (u+1)=-ln(1+sqrt(1-x^2)) +c
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