question_answer

Evaluate \[\int{\frac{\sqrt{{{x}^{2}}+1}[\log |{{x}^{2}}+1|-\,2\,\log |x|]}{{{x}^{4}}}}dx.\]

Answer:

Let \[l=\int{\frac{\sqrt{{{x}^{2}}+1}[\log |{{x}^{2}}+1|-\,2\log |x|}{{{x}^{4}}}dx}\] \[l=\int{\frac{\sqrt{{{x}^{2}}+1}}{{{x}^{4}}}}[\log |{{x}^{2}}+1|-\log |{{x}^{2}}|]dx\]                                     \[[\because \,\,\,mlog\,n=log\,{{n}^{m}}]\] \[=\int{\frac{1}{{{x}^{4}}}\sqrt{{{x}^{2}}\left( 1+\frac{1}{{{x}^{2}}} \right)}\log \left| \frac{{{x}^{2}}+1}{{{x}^{2}}} \right|}\,dx\]                         \[\left[ \because \,\,\log \,m-\log \,n=\log \frac{m}{n} \right]\] \[=\int{\frac{1}{{{x}^{4}}}\cdot x\sqrt{1+\frac{1}{{{x}^{2}}}}}\log \left| 1+\frac{1}{{{x}^{2}}} \right|dx\] \[=\int{\sqrt{1+\frac{1}{{{x}^{2}}}}}\log \left| 1+\frac{1}{{{x}^{2}}} \right|\frac{1}{{{x}^{3}}}\,dx\] Put \[1+\frac{1}{{{x}^{2}}}=t\] \[\Rightarrow \] \[\frac{-\,2}{{{x}^{3}}}dx=dt\] \[\Rightarrow \] \[\frac{1}{{{x}^{3}}}\,dx=-\frac{dt}{2}\] \[\therefore \]      \[l=-\int{\sqrt{\underset{II}{\mathop{t}}\,}\underset{I}{\mathop{\log |t|}}\,\frac{dt}{2}}\] Applying integration by part, we get \[l=\frac{-\,1}{2}\left\{ \frac{2}{3}(log|t|){{t}^{3/2}}-\frac{2}{3}\int{\frac{1}{t}\times {{t}^{3/2}}dt} \right\}\] \[=\frac{-\,1}{2}\left\{ \frac{2}{3}(log|t|){{t}^{3/2}}-\frac{2}{3}\int{{{t}^{1/2}}dt} \right\}\] \[=\frac{-\,1}{2}\left\{ \frac{2}{3}(log|t|){{t}^{3/2}}-\frac{4}{9}{{t}^{3/2}} \right\}+C\] \[=-\frac{1}{3}{{t}^{3/2}}\left\{ \log |t|-\frac{2}{3} \right\}+C\] \[=-\frac{1}{3}{{\left( 1+\frac{1}{{{x}^{2}}} \right)}^{3/2}}\left\{ \log \left| 1+\frac{1}{{{x}^{2}}} \right|-\frac{2}{3} \right\}+C\]                                             \[\left[ \because \,\,\,\,t=1+\frac{1}{{{x}^{2}}} \right]\]