Evaluate the integral: I = integral dx/((x+1) sqrt(1-x))

Evaluate the integral: $$ I = \int \frac{dx}{(x+1)\sqrt{1-x}}. $$

Step 1 · Substitute

Let $u=\sqrt{1-x}$, so $x=1-u^2$, $dx=-2u\,du$, and $x+1=2-u^2$:

$$ I = \int\frac{-2u\,du}{(2-u^2)u} = -2\int\frac{du}{2-u^2}. $$

Step 2 · Apply the standard formula

With $a=\sqrt2$: $\displaystyle\int\frac{du}{a^2-u^2}=\frac1{2a}\ln\left|\frac{a+u}{a-u}\right|$, so:

$$ -2\int\frac{du}{2-u^2} = -\frac1{\sqrt2}\ln\left|\frac{\sqrt2+u}{\sqrt2-u}\right| + C. $$

Step 3 · Back-substitute

$$ I = -\frac1{\sqrt2}\ln\left|\frac{\sqrt2+\sqrt{1-x}}{\sqrt2-\sqrt{1-x}}\right| + C. $$

Summary Table

Additional Notes

• The domain requires $x<1$ (with $x\neq-1$ automatically satisfied since $u\geq0$ there).
• The result was verified by differentiation: writing $f(u)=-\tfrac1{\sqrt2}\ln\left|\tfrac{\sqrt2+u}{\sqrt2-u}\right|$, $\tfrac{df}{du}=-\tfrac2{2-u^2}$, and $\tfrac{du}{dx}=-\tfrac1{2u}$, so $\tfrac{dI}{dx}=\tfrac1{u(2-u^2)}=\tfrac1{\sqrt{1-x}(x+1)}$, matching the integrand exactly.