allery$\int \dfrac{dx}{x(x^2+1)}$ equals
A$\log|x|-\dfrac{1}{2}\log(x^2+1)+C$
B$\log|x|+\dfrac{1}{2}\log(x^2+1)+C$
C$-\log|x|+\dfrac{1}{2}\log(x^2+1)+C$
D$\dfrac{1}{2}\log|x|+\log(x^2+1)+C$
Answer & Solution
Correct answer: A. $\log|x|-\dfrac{1}{2}\log(x^2+1)+C$
1. Write $\dfrac{1}{x(x^2+1)}=\dfrac{A}{x}+\dfrac{Bx+C}{x^2+1}$.
2. Solving: $A=1,\ B=-1,\ C=0$, so integrand $=\dfrac{1}{x}-\dfrac{x}{x^2+1}$.
3. $\int \dfrac{1}{x}\,dx=\log|x|$; $\int \dfrac{x}{x^2+1}\,dx=\dfrac{1}{2}\log(x^2+1)$.
4. Result: $\log|x|-\dfrac{1}{2}\log(x^2+1)+C$. Option C flips the first sign.
_Source: syllabus Mathematics Ch 7 "Integrals", p.322_
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