JEE MAIN - Mathematics (2005 - No. 19)
$$\int {{{\left\{ {{{\left( {\log x - 1} \right)} \over {1 + {{\left( {\log x} \right)}^2}}}} \right\}}^2}\,\,dx} $$ is equal to
$${{\log x} \over {{{\left( {\log x} \right)}^2} + 1}} + C$$
$${x \over {{x^2} + 1}} + C$$
$${{x{e^x}} \over {1 + {x^2}}} + C$$
$${x \over {{{\left( {\log x} \right)}^2} + 1}} + C$$
Explanation
$$\int {{{{{\left( {\log x - 1} \right)}^2}} \over {{{\left( {1 + {{\left( {\log x} \right)}^2}} \right)}^2}}}} dx$$
$$ = \int {{{1 + {{\left( {\log x} \right)}^2} - 2\log x} \over {{{\left[ {1 + {{\left( {\log x} \right)}^2}} \right]}^2}}}} $$
$$ = \int {\left[ {{1 \over {\left( {1 + {{\left( {\log x} \right)}^2}} \right)}} - {{2\log x} \over {{{\left( {1 + {{\left( {\log x} \right)}^2}} \right)}^2}}}} \right]} dx$$
$$ = \int {\left[ {{{{e^t}} \over {1 + {t^2}}} - {{2t\,{e^t}} \over {{{\left( {1 + {t^2}} \right)}^2}}}} \right]} dt$$
put $$\log x = t \Rightarrow dx = {e^t}\,dt$$
$$ = \int {{e^t}} \left[ {{1 \over {1 + {t^2}}} - {{2t} \over {{{\left( {1 + {t^2}} \right)}^2}}}} \right]dt$$
$$\left[ \, \right.$$ which is of the form
$$\left. {\int {{e^x}\left( {f\left( x \right) + f'\left( x \right)dx} \right)} } \right]$$
$$ = {{{e^t}} \over {1 + {t^2}}} + c = {x \over {1 + {{\left( {\log x} \right)}^2}}} + c$$
$$ = \int {{{1 + {{\left( {\log x} \right)}^2} - 2\log x} \over {{{\left[ {1 + {{\left( {\log x} \right)}^2}} \right]}^2}}}} $$
$$ = \int {\left[ {{1 \over {\left( {1 + {{\left( {\log x} \right)}^2}} \right)}} - {{2\log x} \over {{{\left( {1 + {{\left( {\log x} \right)}^2}} \right)}^2}}}} \right]} dx$$
$$ = \int {\left[ {{{{e^t}} \over {1 + {t^2}}} - {{2t\,{e^t}} \over {{{\left( {1 + {t^2}} \right)}^2}}}} \right]} dt$$
put $$\log x = t \Rightarrow dx = {e^t}\,dt$$
$$ = \int {{e^t}} \left[ {{1 \over {1 + {t^2}}} - {{2t} \over {{{\left( {1 + {t^2}} \right)}^2}}}} \right]dt$$
$$\left[ \, \right.$$ which is of the form
$$\left. {\int {{e^x}\left( {f\left( x \right) + f'\left( x \right)dx} \right)} } \right]$$
$$ = {{{e^t}} \over {1 + {t^2}}} + c = {x \over {1 + {{\left( {\log x} \right)}^2}}} + c$$
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