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

$\frac{{x{e^x}}}{{1\,\, + \,\,{x^2}}}\,\, + \,\,c$

$\frac{x}{{{x^2}\,\, + \,\,1}}\,\, + \,\,c$

$\frac{{\log \,\,x}}{{{{\left( {\log \,\,x} \right)}^2} + 1}}\,\, + \,\,c$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

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Analyzing +4 | -1

\int \,\,{\left\{ {\frac{{\log \,\,x - 1}}{{1 + (\log \,\,x){\;^2}}}} \right\}^2}dx

is equal to

View answer and explanation

Correct option: A

\frac{x}{{{{(\log \,\,x)}^2} + \,\,1}} + \,\,c

Option A (

\frac{x}{{{{(\log \,x)}^2} + \,1}} + \,c

) is correct because the derivative of

\frac{x}{{{{(\log \,x)}^2} + \,1}}

is equal to the given integrand.