Evaluate the integral: I = integral dx/(1+ sqrt(1-2x-x²))
Evaluate the integral: $$ I = \int \frac{dx}{1+\sqrt{1-2x-x^2}}. $$
Explanation of Method
Shifting $u=x+1$ centers the quadratic as $2-u^2$; rationalizing by the conjugate $1-\sqrt{2-u^2}$ splits the integral into a rational piece $\int du/(u^2-1)$ and a radical piece $\int\sqrt{2-u^2}/(u^2-1)\,du$, the latter resolved by the trig substitution $u=\sqrt2\sin\theta$.
- Shift $u=x+1$.
- Rationalize by the conjugate.
- Integrate via $u=\sqrt2\sin\theta$.
- Simplify and back-substitute.
Step 1 · Shift
Let $u=x+1$, so $1-2x-x^2=2-u^2$:
$$ I = \int\frac{du}{1+\sqrt{2-u^2}}. $$
Step 2 · Rationalize
Multiplying by $\dfrac{1-\sqrt{2-u^2}}{1-\sqrt{2-u^2}}$, the denominator becomes $1-(2-u^2)=u^2-1$:
$$ I = \int\frac{du}{u^2-1} - \int\frac{\sqrt{2-u^2}}{u^2-1}\,du. $$
Step 3 · Trig substitution on the second piece
Let $u=\sqrt2\sin\theta$: this reduces $\int\sqrt{2-u^2}/(u^2-1)\,du$ to a standard rational-trig form, and after simplification (combined with the first piece's $\tfrac12\ln|(u-1)/(u+1)|$), the two logarithmic pieces collapse via the identity $\sec2\theta+\tan2\theta=(1+u\sqrt{2-u^2})/(1-u^2)$:
$$ I = \arcsin\left(\frac u{\sqrt2}\right) + \frac12\ln\left|1+u\sqrt{2-u^2}\right| - \ln|1+u| + C. $$
Step 4 · Back-substitute
With $u=x+1$, $2-u^2=1-2x-x^2$:
Summary Table
| Piece | Antiderivative |
|---|---|
| $1/(u^2-1)$ | $\tfrac12\ln|(u-1)/(u+1)|$ |
| $-\sqrt{2-u^2}/(u^2-1)$ | combines with above into $\arcsin(u/\sqrt2)+\tfrac12\ln|1+u\sqrt{2-u^2}|-\ln|1+u|$ |
Additional Notes
- Major rewrite: the source's Steps 6–7 carried out a convoluted trig reduction and then produced a final "simplified form", $\arcsin\left((x+1)/\sqrt2\right)-\sqrt{1-2x-x^2}$, with the note "this simplified form can be verified by differentiation" — without actually verifying it. Testing it at $x=-\tfrac12$ gives a derivative of $\approx1.134$, while the true integrand there is $\approx0.4305$. The boxed result below was obtained via an explicit rationalization and trig-substitution derivation and was verified independently.
- The domain requires $1-2x-x^2>0$, i.e. $-1-\sqrt2<x<-1+\sqrt2$.
- The result was verified by differentiation at $x=-1,\,-0.7,\,-1.3$: the derivative matches the integrand $1/\left[1+\sqrt{1-2x-x^2}\right]$ exactly at all three points.
Result
The solution to the integral is: