allery$\displaystyle\int \log x\,dx$ equals:
A$\dfrac{(\log x)^2}{2} + C$
B$x \log x + C$
C$\dfrac{1}{x} + C$
D$x \log x - x + C$
Answer & Solution
Correct answer: D. $x \log x - x + C$
Apply integration by parts with $u = \log x$ and $dv = dx$ (treat $1$ as the second function).
Then $du = \dfrac{1}{x}\,dx$, $v = x$, giving $\int \log x\,dx = x \log x - \int x \cdot \dfrac{1}{x}\,dx = x \log x - x + C$.
Keep practising Unit 6A: Indefinite Integrals & Fundamental Theorem
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