\(\displaystyle\int \left\lbrace \dfrac{(\log x - 1)}{ 1 + (\log x)^2}\right\rbrace^2dx\) is equal to

\(\rm \dfrac{\log x}{(\log x)^2 + 1}+C\)

\(\rm \dfrac{x}{(\log x)^2 +1}+C\)

\(\displaystyle\int \left\lbrace \dfrac{(\log x - 1)}{ 1 + (\log x)^2}

1.	\(\displaystyle\int \left\lbrace \dfrac{(\log x - 1)}{ 1 + (\log x)^2}\right\rbrace^2dx\) is equal to
A.	\(\rm \dfrac{xe^x}{1+x^2}+C\)
B.	\(\rm \dfrac{x}{(\log x)^2 +1}+C\)
C.	\(\rm \dfrac{\log x}{(\log x)^2 + 1}+C\)
D.	\(\rm \dfrac{x}{x^2 + 1}+C\)
Answer» C. \(\rm \dfrac{\log x}{(\log x)^2 + 1}+C\)

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\(\displaystyle\int \left\lbrace \dfrac{(\log x - 1)}{ 1 + (\log x)^2}\right\rbrace^2dx\) is equal to

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