. $\cfrac{{\sqrt {{x^2} + 1} \left[ {\log \left( {{x^2} + 1} \right) - 2\log x} \right]}}{{{x^4}}}$

Let $I = \int {\sqrt {{x^2} + 1} \cfrac{{\left[ {\log \left( {{x^2} + 1} \right) - 2\log x} \right]}}{{{x^4}}}} dx$

$= \int {\sqrt {\cfrac{{{x^2} + 1}}{{{x^2}}}} \left[ {\log \left( {\cfrac{{{x^2} + 1}}{{{x^2}}}} \right)} \right]} \cfrac{{dx}}{{{x^3}}} = \int {\sqrt {1 + \cfrac{1}{{{x^2}}}} \left[ {\log \left( {1 + \cfrac{1}{{{x^2}}}} \right)} \right]} \cfrac{{dx}}{{{x^3}}}$

Put $\cfrac{1}{{{x^2}}} = t$ $\Rightarrow$ $- 2{x^{ - 3}}dx = dt$

$\Rightarrow$ $- \cfrac{2}{{{x^3}}}dx = dt$

$\therefore$ $I = \cfrac{{ - 1}}{2}\int {\sqrt {1 + t} \log \left( {1 + t} \right)dt}$

$\therefore$ $I = \cfrac{{ - 1}}{2}\left[ {\log \left( {1 + t} \right) \cdot \cfrac{{{{\left( {1 + t} \right)}^{3/2}}}}{{3/2}} - \int {\cfrac{1}{{1 + t}}\cfrac{{{{\left( {1 + t} \right)}^{3/2}}}}{{3/2}}} dt} \right]$

$= - \cfrac{1}{2}\left[ {\cfrac{2}{3}{{\left( {1 + t} \right)}^{3/2}}\log \left( {1 + t} \right) - \cfrac{2}{3}\int {{{\left( {1 + t} \right)}^{1/2}}dt} } \right]$

$= - \cfrac{1}{2}\left[ {\cfrac{2}{3}{{\left( {1 + t} \right)}^{3/2}}\log \left( {1 + t} \right) - \cfrac{2}{3}\cfrac{{{{\left( {1 + t} \right)}^{3/2}}}}{{3\,\,3/2}}} \right] + C$

$= - \cfrac{1}{2}\left[ {\cfrac{2}{3}{{\left( {1 + t} \right)}^{3/2}}\log \left( {1 + t} \right) - \cfrac{4}{9}{{\left( {1 + t} \right)}^{3/2}}} \right] + C$

$= - \cfrac{1}{3}{\left( {1 + t} \right)^{3/2}}\log \left( {1 + t} \right) + \cfrac{2}{9}{\left( {1 + t} \right)^{3/2}} + C$

$= - \cfrac{1}{3}{\left( {1 + \cfrac{1}{{{x^2}}}} \right)^{3/2}}\log \left( {1 + \cfrac{1}{{{x^2}}}} \right) + \cfrac{2}{9}{\left( {1 + \cfrac{1}{{{x^2}}}} \right)^{3/2}} + C$

$= - \cfrac{1}{3}{\left( {1 + \cfrac{1}{{{x^2}}}} \right)^{3/2}}\left[ {\left( {\log \left( {1 + \cfrac{1}{{{x^2}}}} \right) - \cfrac{2}{3}} \right)} \right] + C$

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