Write the integral in terms of $u$ and $d u$. Then evaluate.
\[
\int(8 x+29)^{-2} d x, \quad u=8 x+29
\]
(Use symbolic notation and fractions where needed.)
\[
\int(8 x+29)^{-2} d x=
\]

Step 1 ：Given that \(u = 8x + 29\), we can differentiate both sides with respect to \(x\) to find \(du\). So, \(du = 8dx\).

Step 2 ：We can solve this for \(dx\) to get \(dx = \frac{1}{8}du\).

Step 3 ：Now, we can substitute \(u\) and \(dx\) into the integral: \[\int(8 x+29)^{-2} d x = \int u^{-2} \cdot \frac{1}{8} du\]

Step 4 ：This simplifies to: \[\frac{1}{8} \int u^{-2} du\]

Step 5 ：Now, we can evaluate the integral. The antiderivative of \(u^{-2}\) is \(-u^{-1}\), so: \[\frac{1}{8} \int u^{-2} du = -\frac{1}{8}u^{-1} + C\]

Step 6 ：Finally, we substitute \(u = 8x + 29\) back into the equation to get the answer in terms of \(x\): \[-\frac{1}{8}(8x + 29)^{-1} + C\]

Step 7 ：So, the integral of \((8x + 29)^{-2} dx\) is \(\boxed{-\frac{1}{8}(8x + 29)^{-1} + C}\)