The integral provided in the image is:

$\int_{-1}^0 \frac{x \, dx}{\sqrt{1 - x^2}}$

Let's solve it step-by-step:

Step 1: Analyze the Integral

The denominator $\sqrt{1 - x^2}$ resembles the derivative of $\arcsin(x)$, but because the numerator has $x$, substitution will simplify this integral.

Step 2: Substitution

Let $u = 1 - x^2$, so: $du = -2x \, dx \quad \text{or} \quad x \, dx = -\frac{1}{2} du.$

When $x = -1$: $u = 1 - (-1)^2 = 0.$

When $x = 0$: $u = 1 - 0^2 = 1.$

The limits of integration change from $x \in [-1, 0]$ to $u \in [0, 1]$.

Step 3: Rewrite the Integral

Substituting these into the integral: $\int_{-1}^0 \frac{x \, dx}{\sqrt{1 - x^2}} = \int_{0}^1 \frac{-\frac{1}{2} \, du}{\sqrt{u}} = -\frac{1}{2} \int_{0}^1 u^{-1/2} \, du.$

Step 4: Solve the Integral

The integral of $u^{-1/2}$ is: $\int u^{-1/2} \, du = 2u^{1/2}.$

Thus: $-\frac{1}{2} \int u^{-1/2} \, du = -\frac{1}{2} \cdot 2u^{1/2} = -u^{1/2}.$

Step 5: Apply Limits of Integration

Now substitute the limits $u = 0$ and $u = 1$: $\left[ -u^{1/2} \right]_0^1 = -1^{1/2} + 0^{1/2} = -1 + 0 = -1.$

Final Answer:

The value of the integral is: $\boxed{-1}.$

Would you like a deeper explanation or have any questions? Here are some related questions to explore:

1. What happens if the limits of integration are swapped?
2. Why is substitution helpful for solving this integral?
3. How would the integral change if the numerator was $x^2$?
4. What if the integrand involved $\sqrt{1 - x^2}$ in the numerator?
5. How does this integral relate to trigonometric functions like $\arcsin(x)$?

Tip: Always check if substitution can simplify an integral when you notice a composite function.