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

Evaluate the integral: $$ I = \int \frac{dx}{(1+x)\sqrt{1-x-x^2}}. $$

Step 1 · Substitute

Let $t=\dfrac1{1+x}$, so $x=\dfrac{1-t}t$ and $dx=-\dfrac{dt}{t^2}$. A direct computation gives:

$$ 1-x-x^2 = \frac{t^2+t-1}{t^2}. $$

Step 2 · Substitute into the integral

For $t>0$ (i.e. $1+x>0$):

$$ I = \int\frac{-dt/t^2}{\frac1t\cdot\frac{\sqrt{t^2+t-1}}t} = -\int\frac{dt}{\sqrt{t^2+t-1}}. $$

Step 3 · Complete the square and integrate

$$ t^2+t-1=\left(t+\frac12\right)^2-\frac54 \quad\Rightarrow\quad -\int\frac{dt}{\sqrt{(t+1/2)^2-5/4}} = -\ln\left|t+\frac12+\sqrt{t^2+t-1}\right| + C. $$

Step 4 · Back-substitute

Since $t+\tfrac12=\dfrac{3+x}{2(1+x)}$ and $\sqrt{t^2+t-1}=\dfrac{\sqrt{1-x-x^2}}{1+x}$:

$$ I = -\ln\left|\frac{(3+x)+2\sqrt{1-x-x^2}}{2(1+x)}\right| + C = \ln\left|1+x\right| - \ln\left|(3+x)+2\sqrt{1-x-x^2}\right| + C. $$

Summary Table

Additional Notes

• Major rewrite: the source's approach chained a trigonometric substitution with a Weierstrass ($t=\tan(\theta/2)$) substitution, producing a final answer nested three substitutions deep — a logarithm of a ratio involving $\tan\left(\tfrac12\arcsin\left(\tfrac{2x+1}{\sqrt5}\right)\right)$ — which the source itself acknowledged was "quite complex" without verifying it. The reciprocal substitution $t=1/(1+x)$ above reaches a clean closed form directly, since it turns the linear factor $(1+x)$ into a constant.
• The domain requires $1-x-x^2>0$ and $x>-1$, i.e. $-1<x<\tfrac{-1+\sqrt5}2\approx0.618$.
• The result was verified by differentiation at $x=0$: the derivative evaluates to $1$, matching the integrand $1/[(1)\sqrt1]=1$ there.