91Ó°ÊÓ

To rewrite the expression inside the square root, we complete the square: $$1-2x-x^2 = -(x^2+2x-1)$$ Now we complete the square for the quadratic inside the parentheses. We take half of the coefficient of the linear term (which is 1) and square it. Then, we add and subtract this number to the quadratic: $$-(x^2+2x+\color{red}{1} -\color{red}{1}-1) = -(x+1)^2+1$$ Now, we can rewrite the integral as: $$ \int \frac{x^{2}}{\sqrt{1-(x+1)^{2}}} dx $$

In this problem, we will use a trigonometric substitution to simplify the expression inside the square root. To do that, we will use the identity \(\sin^2{θ} + \cos^2{θ} = 1\). We also want to ensure that the \(dx\) term from the numerator can be replaced with the corresponding \(dθ\) term. The substitution we will use is: $$x + 1 = \cos{θ}$$ Thus, \(dx = -\sin{θ} dθ\). Now, we substitute these expressions back into the integral: $$\int \frac{x^{2}}{\sqrt{1-(x+1)^{2}}} dx = \int \frac{(\cos{θ} - 1)^{2}}{\sqrt{1-\cos^{2}{θ}}} (-\sin{θ}) dθ$$

We can simplify the expression under the square root using the identity \(\sin^2{θ}+\cos^2{θ}=1\): $$\int \frac{(\cos{θ} - 1)^{2}}{\sqrt{1-\cos^{2}{θ}}} (-\sin{θ}) dθ = \int \frac{(\cos{θ} - 1)^{2}}{\sin{θ}} (-\sin{θ}) dθ$$ The \(\sin{θ}\) terms cancel out, leaving us with: $$-\int(\cos{θ}-1)^2 dθ$$ Now, we expand the square and integrate term by term: $$-\int(\cos^2{θ} - 2\cos{θ} +1) dθ = -\int \cos^2{θ} dθ +2\int \cos{θ} dθ -\int 1 dθ$$ We can evaluate these integrals separately: $$-\int \cos^2{θ} dθ = -\frac{1}{2} \int (1 + \cos{2θ}) dθ = -\frac{1}{2} (θ+\frac{1}{2}\sin{2θ}) + C_1$$ $$2\int \cos{θ} dθ = 2\sin{θ} + C_2$$ $$-\int 1 dθ = -θ + C_3$$ Adding the results of these integrals back together, we get: $$-\frac{1}{2} (θ+\frac{1}{2}\sin{2θ}) + 2\sin{θ} -θ + C = -\frac{1}{2}θ-\frac{1}{4}\sin{2θ}+2\sin{θ}-θ+C$$

Now we need to express our result in terms of \(x\) instead of \(θ\). Using our substitution, we can find the value of \(θ\): $$x+1=\cos{θ} \Rightarrow θ = \arccos(x+1)$$ We can also use the right triangle: $$\sin{θ} = \sqrt{1-\cos^2{θ}} = \sqrt{1-(x+1)^2} = -\sqrt{-(x+1)^2+1} = \sqrt{1-x^2-2x}$$ Then, we can find the value of \(\sin{2θ}\): $$\sin{2θ} = 2\sin{θ}\cos{θ} = 2\sqrt{1-x^2-2x}(x+1)$$ Finally, we substitute these values back into our result: $$-\frac{1}{2}\arccos(x+1)-\frac{1}{4}(2\sqrt{1-x^2-2x}(x+1))+2\sqrt{1-x^2-2x}-\arccos(x+1)+C$$ So the final answer is: $$-\frac{3}{2}\arccos(x+1)-\frac{1}{2}(x+1)\sqrt{1-x^2-2x}+2\sqrt{1-x^2-2x}+C$$