$int(10 - 0.8v)^{-1}dv$

Explanation:

Step1: Use substitution method

Let $u = 10 - 0.8v$, then $du=-0.8dv$, and $dv=-\frac{1}{0.8}du = -\frac{5}{4}du$.

Step2: Rewrite the integral

The original integral $\int(10 - 0.8v)^{-1}dv$ becomes $\int u^{-1}\left(-\frac{5}{4}\right)du=-\frac{5}{4}\int\frac{1}{u}du$.

Step3: Integrate $\frac{1}{u}$

We know that $\int\frac{1}{u}du=\ln|u| + C$. So $-\frac{5}{4}\int\frac{1}{u}du=-\frac{5}{4}\ln|u|+C$.

Step4: Substitute back $u$

Substitute $u = 10 - 0.8v$ back, we get $-\frac{5}{4}\ln|10 - 0.8v|+C$.

Answer:

$-\frac{5}{4}\ln|10 - 0.8v|+C$