$\int \log x\,dx$ equals
A$x\log x+x+C$
B$\dfrac{1}{x}+C$
C$x\log x-x+C$
D$\log x-x+C$
Answer & Solution
Correct answer: C. $x\log x-x+C$
1. Take $u=\log x$ (first), $dv=1\,dx$, so $v=x$.
2. By parts: $x\log x-\int x\cdot\dfrac{1}{x}\,dx$.
3. $=x\log x-\int 1\,dx=x\log x-x+C$.
4. Differentiating: $\log x+1-1=\log x$, confirming C.
_Source: NCERT Class 12 Mathematics Ch 7 "Integrals", p.325_
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