Evaluate the integral: I = integral x/(sqrt(3-2x-x²)) dx
Evaluate the integral: $$ I = \int \frac{x}{\sqrt{3-2x-x^2}}\,dx. $$
Explanation of Method
Completing the square and shifting to $u=x+1$ turns the numerator into $u-1$, splitting the integral into a derivative-matching square-root piece and a standard arcsine piece.
- Complete the square for $3-2x-x^2$.
- Substitute $u=x+1$.
- Split the numerator $u-1$ and integrate each piece.
- Substitute back.
Step 1 · Complete the square
$$ 3-2x-x^2 = 4-(x+1)^2. $$
Step 2 · Substitute
Let $u=x+1$, so $x=u-1$, $dx=du$:
$$ I = \int \frac{u-1}{\sqrt{4-u^2}}\,du = \int \frac{u\,du}{\sqrt{4-u^2}} - \int \frac{du}{\sqrt{4-u^2}}. $$
Step 3 · Integrate each piece
For the first piece, let $w=4-u^2$, $dw=-2u\,du$:
$$ \int \frac{u\,du}{\sqrt{4-u^2}} = -\sqrt{4-u^2}. $$
For the second piece, using the standard arcsine formula:
$$ \int \frac{du}{\sqrt{4-u^2}} = \arcsin\frac u2. $$
Step 4 · Combine and substitute back
$$ I = -\sqrt{4-u^2} - \arcsin\frac u2 + C. $$
With $u=x+1$ and $4-u^2=3-2x-x^2$:
$$ I = -\sqrt{3-2x-x^2} - \arcsin\left(\frac{x+1}{2}\right) + C. $$
Summary Table
| Piece | Antiderivative |
|---|---|
| $u/\sqrt{4-u^2}$ | $-\sqrt{4-u^2}$ |
| $1/\sqrt{4-u^2}$ | $\arcsin(u/2)$ |
Additional Notes
- The domain requires $3-2x-x^2\geq0$, i.e. $-3\leq x\leq1$.
- This is the same shift-and-split technique as Problem 215, but with a bounded $a^2-u^2$ radical (arcsine) rather than an unbounded $u^2+a^2$ radical (arctangent).
- The result can be checked by differentiating $-\sqrt{3-2x-x^2}-\arcsin\!\left(\tfrac{x+1}2\right)$: the first term gives $\tfrac{1+x}{\sqrt{3-2x-x^2}}$, and the second gives $-\tfrac{1}{\sqrt{3-2x-x^2}}$ (using $4-(x+1)^2=3-2x-x^2$); summing gives $\tfrac{x}{\sqrt{3-2x-x^2}}$, matching the integrand.
Result
The solution to the integral is: