integral of -1.436/r
You asked:
Evaluate the integral: \(\int{\frac{-1.436}{r}}dr\)
MathBot Answer:
Antiderivative
\[\int{\frac{-1.436}{r}}dr = - \frac{359 \log{\left(r \right)}}{250} + C\]
integral of -1.436/r
You asked:
MathBot Answer:
\[\int{\frac{-1.436}{r}}dr = - \frac{359 \log{\left(r \right)}}{250} + C\]