The integral $$\int \frac{x \ dx}{2-x^2+\sqrt{2-x^2}}$$ equals :

We are given the integral: $$\int \frac{x dx}{2 - x^2 + \sqrt{2 - x^2}}$$

To solve this, we notice the expression $$\sqrt{2 - x^2}$$ in the denominator and the term $$2 - x^2$$. This suggests a substitution. Let $$t = \sqrt{2 - x^2}$$. Then, squaring both sides, we get $$t^2 = 2 - x^2$$.

Differentiate both sides with respect to $$x$$:

$$\frac{d}{dx}(t^2) = \frac{d}{dx}(2 - x^2)$$

$$2t \frac{dt}{dx} = -2x$$

Solving for $$dx$$, we rearrange to get $$x dx = -t dt$$. This is because:

$$2t dt = -2x dx \implies t dt = -x dx \implies x dx = -t dt$$

Now, substitute into the integral. Replace $$\sqrt{2 - x^2}$$ with $$t$$ and $$2 - x^2$$ with $$t^2$$, and replace $$x dx$$ with $$-t dt$$:

$$\int \frac{x dx}{2 - x^2 + \sqrt{2 - x^2}} = \int \frac{-t dt}{t^2 + t}$$

Simplify the denominator: $$t^2 + t = t(t + 1)$$. So the integral becomes:

$$\int \frac{-t dt}{t(t + 1)}$$

Assuming $$t \neq 0$$, we can cancel $$t$$ in the numerator and denominator:

$$\int \frac{-t}{t(t + 1)} dt = \int \frac{-1}{t + 1} dt$$

This simplifies to:

$$-\int \frac{1}{t + 1} dt$$

The integral of $$\frac{1}{t + 1}$$ is $$\log|t + 1|$$:

$$-\int \frac{1}{t + 1} dt = -\log|t + 1| + c$$

Now, substitute back $$t = \sqrt{2 - x^2}$$:

$$-\log|\sqrt{2 - x^2} + 1| + c$$

Since $$\sqrt{2 - x^2} \geq 0$$ and adding 1 makes it always positive (greater than or equal to 1), the absolute value can be written without changing the expression. However, to match the options, we keep the absolute value:

$$-\log\left|1 + \sqrt{2 - x^2}\right| + c$$

Comparing this result with the given options:

• Option A: $$\log\left|1 + \sqrt{2 + x^2}\right| + c$$ — has $$2 + x^2$$, not matching.
• Option B: $$-\log\left|1 + \sqrt{2 - x^2}\right| + c$$ — matches exactly.
• Option C: $$-x\log\left|1 - \sqrt{2 - x^2}\right| + c$$ — has an extra $$x$$ and a minus sign inside the log, not matching.
• Option D: $$x\log\left|1 - \sqrt{2 + x^2}\right| + c$$ — has $$x$$ and $$2 + x^2$$, not matching.

Hence, the correct answer is Option B.