Question

If u and v are differentiable functions of x, then prove that \( \int u \, dv = uv - \int v \, du \). Hence evaluate \( \int \log x \, dx \).

Hint:

Integration by parts: \( \int u dv = uv - \int v du \); choose u as logarithmic, dv as dx.

Answer 1

Proof: Differentiate both sides:
\[ \frac{d}{dx} (uv) = u \frac{dv}{dx} + v \frac{du}{dx}. \] Integrate: \( uv = \int u \frac{dv}{dx} dx + \int v \frac{du}{dx} dx \).
\[ uv = \int u \, dv + \int v \, du $\Rightarrow$ \int u \, dv = uv - \int v \, du. \] Hence: Let \( u = \log x \), \( dv = dx \).
\[ du = \frac{1}{x} dx, v = x. \] \[ \int \log x \, dx = x \log x - \int x \cdot \frac{1}{x} dx = x \log x - \int 1 \, dx = x \log x - x + c. \] Answer: Proved; \( x \log x - x + c \).