Question

If $ \int e^{-3 \log x} \, dx = f(x) + C $, then $ f(x) $ is:

• A. $ e^{-3 \log x} $
• B. $ e $
• C. $ \frac{-1}{2x^2} $
• D. $ \frac{-1}{4x^4} $

Hint:

When solving integrals of the form $ e^{a \log x} $, use the logarithmic identity $ e^{\log x} = x $. This allows you to simplify the expression before integrating.

Answer 1

The Correct Option is C

We are given the integral: \[ \int e^{-3 \log x} \, dx = f(x) + C \] First, simplify the expression inside the integral: \[ e^{-3 \log x} = \left( e^{\log x} \right)^{-3} = x^{-3} \] So, the integral becomes: \[ \int x^{-3} \, dx \] Step 1: Integrate $x^{-3}$ with respect to $x$: \[ \int x^{-3} \, dx = \frac{x^{-2}}{-2} = \frac{-1}{2x^2} \] Thus, we have: \[ f(x) = \frac{-1}{2x^2} \] Therefore, the correct answer is (C) $ \frac{-1}{2x^2} $.