Find the indefinite integral. (Remember the constant of integration.) \[ \int \frac{7}{81+(5-x)^{2}} d x \]

To solve the integral \[ \int \frac{7}{81 + (5 - x)^2} \, dx, \] we can use a substitution to simplify it. Let \( u = 5 - x \), which gives us \( du = -dx \), so \( dx = -du \). The bounds of the integral change accordingly, but since we're finding an indefinite integral, we won't worry about bounds. Substituting, we have: \[ \int \frac{7}{81 + u^2} (-du) = -7 \int \frac{1}{81 + u^2} \, du. \] The integral \[ \int \frac{1}{a^2 + u^2} \, du = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C, \] is known, where \( a = 9 \) (since \( 81 = 9^2 \)). Thus, we have: \[ -7 \cdot \frac{1}{9} \tan^{-1}\left(\frac{u}{9}\right) + C = -\frac{7}{9} \tan^{-1}\left(\frac{5 - x}{9}\right) + C. \] Finally, the result of the indefinite integral is \[ -\frac{7}{9} \tan^{-1}\left(\frac{5 - x}{9}\right) + C. \]