Question
$$\int {{{\left\{ {\frac{{\left( {\log \,x - 1} \right)}}{{1 + {{\left( {\log \,x} \right)}^2}}}} \right\}}^2}dx} $$ is equal to-
A.
$$\frac{{\log \,x}}{{{{\left( {\log \,x} \right)}^2} + 1}} + C$$
B.
$$\frac{x}{{{x^2} + 1}} + C$$
C.
$$\frac{{x{e^x}}}{{1 + {x^2}}} + C$$
D.
$$\frac{x}{{{{\left( {\log \,x} \right)}^2} + 1}} + C$$
Answer :
$$\frac{x}{{{{\left( {\log \,x} \right)}^2} + 1}} + C$$
Solution :
$$\eqalign{
& \int {\frac{{{{\left( {\log \,x - 1} \right)}^2}}}{{{{\left( {1 + {{\left( {\log \,x} \right)}^2}} \right)}^2}}}dx} \cr
& = \int {\frac{{1 + {{\left( {\log \,x} \right)}^2} - 2\,\log \,x}}{{{{\left[ {1 + {{\left( {\log \,x} \right)}^2}} \right]}^2}}}} dx \cr
& = \int {\left[ {\frac{1}{{\left( {1 + {{\left( {\log \,x} \right)}^2}} \right)}} - \frac{{2\,\log \,x}}{{{{\left( {1 + {{\left( {\log \,x} \right)}^2}} \right)}^2}}}} \right]} dx \cr
& = \int {\left[ {\frac{{{e^t}}}{{1 + {t^2}}} - \frac{{2t\,{e^t}}}{{{{\left( {1 + {t^2}} \right)}^2}}}} \right]} dt\,\, \cr
& {\text{put }}\log \,x = t\,\, \Rightarrow dx = {e^{t\,}}dt \cr
& = \int {{e^t}} \left[ {\frac{1}{{1 + {t^2}}} - \frac{{2t}}{{{{\left( {1 + {t^2}} \right)}^2}}}} \right]dt \cr
& \left[ {{\text{which is of the form }}\int {{e^x}\left( {f\left( x \right) + f'\left( x \right)} \right)dx} } \right] \cr
& = \frac{{{e^t}}}{{1 + {t^2}}} + C\,\,\,\, = \frac{x}{{1 + {{\left( {\log \,x} \right)}^2}}} + C \cr} $$