integral of -1.436/r

asked by guest
on Nov 16, 2024 at 7:13 pm



You asked:

Evaluate the integral: ∫−1.436rdr\int{\frac{-1.436}{r}}dr

MathBot Answer:

Antiderivative

∫−1.436rdr=−359log⁡(r)250+C\int{\frac{-1.436}{r}}dr = - \frac{359 \log{\left(r \right)}}{250} + C

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