alleryUsing the substitution $t = 1 + x^2$, $\displaystyle\int \dfrac{x}{1 + x^2}\,dx$ equals:
A$\dfrac{1}{2}\log(1 + x^2) + C$
B$\dfrac{1}{2}\tan^{-1}(x^2) + C$
C$\log(1 + x^2) + C$
D$\tan^{-1} x + C$
Answer & Solution
Correct answer: A. $\dfrac{1}{2}\log(1 + x^2) + C$
With $t = 1 + x^2$, $dt = 2x\,dx$, so $\int \dfrac{x}{1+x^2}\,dx = \dfrac{1}{2}\int \dfrac{dt}{t} = \dfrac{1}{2}\log|t| + C = \dfrac{1}{2}\log(1 + x^2) + C$.
Note the $1/2$ factor, which trips up answers like option B.
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