allery$\int \dfrac{(\log x)^2}{x}\,dx$ equals
A$\dfrac{1}{x}(\log x)^3+C$
B$2\log x+C$
C$\dfrac{(\log x)^2}{2}+C$
D$\dfrac{(\log x)^3}{3}+C$
Answer & Solution
Correct answer: D. $\dfrac{(\log x)^3}{3}+C$
1. Put $t=\log x$, so $dt=\dfrac{1}{x}dx$.
2. Integral $=\int t^2\,dt=\dfrac{t^3}{3}$.
3. Back-substitute: $\dfrac{(\log x)^3}{3}+C$.
4. Differentiating: $\dfrac{3(\log x)^2}{3}\cdot\dfrac{1}{x}=\dfrac{(\log x)^2}{x}$, confirming A.
_Source: syllabus Mathematics Ch 7 "Integrals", p.302_
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