allery$\int \dfrac{2x}{1+x^2}\,dx$ equals
A$\tan^{-1}x+C$
B$2\log(1+x^2)+C$
C$\dfrac{1}{1+x^2}+C$
D$\log(1+x^2)+C$
Answer & Solution
Correct answer: D. $\log(1+x^2)+C$
1. Put $t=1+x^2$, so $dt=2x\,dx$ (the numerator).
2. Integral $=\int \dfrac{dt}{t}=\log|t|$.
3. Since $1+x^2>0$: $\log(1+x^2)+C$.
4. Option A would need numerator $1$, not $2x$.
_Source: syllabus Mathematics Ch 7 "Integrals", p.302_
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