Identify the integral.

The integral to evaluate is:

$$\int \frac{x}{49-x^2} dx$$

Let $u = 49 - x^2$.
Differentiate $u$ with respect to $x$:

$$\frac{du}{dx} = -2x$$

Rearrange to express $x \, dx$ in terms of $du$:

$$du = -2x \, dx$$ $$x \, dx = -\frac{1}{2} du$$

Substitute $u$ for $49-x^2$ and $-\frac{1}{2} du$ for $x \, dx$:

$$\int \frac{1}{u} \left(-\frac{1}{2}\right) du$$ $$= -\frac{1}{2} \int \frac{1}{u} du$$

The integral of $\frac{1}{u}$ is $\ln|u|$.

$$-\frac{1}{2} \int \frac{1}{u} du = -\frac{1}{2} \ln|u| + C$$

Replace $u$ with $49 - x^2$ to get the final answer in terms of $x$:

$$-\frac{1}{2} \ln|49 - x^2| + C$$