∫2xx2+1dx Put x2+1=f(x)2xdx=f′(x)=∫f′(x)f(x)dx=ln⁡|f(x)|+c=ln⁡x2+1+c

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$\int \frac{2 x}{x^{2} + 1} d x =$

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$\ln |x + 2| + c$

$- \ln |2 x + 1| + c$

$\ln |2 x - 1| + c$

$\ln |x^{2} + 1| + c$

answer is D.

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Detailed Solution

$\begin{array}{l} \int \frac{2 x}{x^{2} + 1} d x \\ Put x^{2} + 1 = f (x) \\ 2 x d x = f^{'} (x) \\ = \int \frac{f^{'} (x)}{f (x)} d x \\ = \ln | f (x) | + c \\ = \ln |x^{2} + 1| + c \end{array}$

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