Question

The value of \( \int \log x \, dx = x \log x + x + c \).

Hint:

The integral of \( \log x \) is a standard result that is frequently used. It's helpful to remember it: \( \int \log x \, dx = x \log x - x + c \).

Answer 1

Step 1: Use integration by parts, where \( \int u \, dv = uv - \int v \, du \). Let \( u = \log x \) and \( dv = dx \).
Step 2: Differentiate \( u \) and integrate \( dv \). We get \( du = \frac{1}{x} dx \) and \( v = x \).
Step 3: Substitute these into the integration by parts formula: \[ \int \log x \, dx = (\log x)(x) - \int x \left(\frac{1}{x}\right) dx = x \log x - \int 1 \, dx = x \log x - x + c. \] The given statement has a positive sign before the \(x\), so it is false.