$\int {{{\left\{ {{{\left( {\log x - 1} \right)} \over {1 + {{\left( {\log x} \right)}^2}}}} \right\}}^2}\,\,dx} $ is equal to

Question

${{\log x} \over {{{\left( {\log x} \right)}^2} + 1}} + C$

${x \over {{x^2} + 1}} + C$

${{x{e^x}} \over {1 + {x^2}}} + C$

${x \over {{{\left( {\log x} \right)}^2} + 1}} + C$

Please login to view the detailed solution steps...