question_answer

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

OR

Evaluate \[\int{x\cdot {{(\log \,\,x)}^{2}}\,dx.}\]

Answer:

Let \[l=\int{\frac{1+{{x}^{2}}}{1+{{x}^{4}}}\,dx=\int{\frac{\frac{1}{{{x}^{2}}}+\frac{{{x}^{2}}}{{{x}^{2}}}}{\frac{1}{{{x}^{2}}}+\frac{{{x}^{4}}}{{{x}^{2}}}}\,dx}}\]

[divide numerator and denominator by \[{{x}^{2}}\]]

\[=\int{\frac{1+\frac{1}{{{x}^{2}}}}{{{x}^{2}}+\frac{1}{{{x}^{2}}}}\,dx}\]

\[=\int{\frac{1+\frac{1}{{{x}^{2}}}}{\left( {{x}^{2}}+\frac{1}{{{x}^{2}}}-2+2 \right)}\,dx}\] \[=\int{\frac{1+\frac{1}{{{x}^{2}}}}{{{\left( x-\frac{1}{{{x}^{2}}} \right)}^{2}}+2}\,dx}\]

Put \[x-\frac{1}{x}=t\] \[\Rightarrow \] \[\left( 1+\frac{1}{{{x}^{2}}} \right)\,dx=dt\]

\[\therefore \] \[l=\int{\frac{dt}{{{t}^{2}}+2}=\int{\frac{dt}{{{t}^{2}}+{{(\sqrt{2})}^{2}}}}}=\frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \frac{t}{\sqrt{2}} \right)\]

\[\left[ \because \,\,\,\int{\frac{1}{{{a}^{2}}+{{x}^{2}}}\,dx=\frac{1}{a}{{\tan }^{-1}}\frac{x}{a}} \right]\]

\[=\frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \frac{x-\frac{1}{x}}{\sqrt{2}} \right)\]                \[\left[ \because \,\,\,t=x-\frac{1}{x} \right]\]

\[=\frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \frac{{{x}^{2}}-1}{\sqrt{2x}} \right)+C\]

OR

Let \[l=\int{\underset{II}{\mathop{x}}\,}\cdot {{(\underset{I}{\mathop{log}}\,)}^{2}}dx\]

Applying integration by parts, we get

\[l={{(\log \,x)}^{2}}\int{x\,dx-\int{\left[ \frac{d}{dx}{{(\log \,x)}^{2}}\cdot \int{x\,dx} \right]}}\,dx\]

\[={{(\log \,x)}^{2}}\cdot \frac{{{x}^{2}}}{2}-\int{\left[ \frac{2\log \,x}{x}\times \frac{{{x}^{2}}}{2} \right]}\,dx\]

\[=\frac{{{x}^{2}}}{2}{{(\log \,x)}^{2}}-\int{\underset{II}{\mathop{x}}\,\underset{I}{\mathop{\log }}\,}\,x\,dx\]

Again, applying integration by parts, we get

\[l=\frac{{{x}^{2}}}{2}{{(\log \,x)}^{2}}-\]

\[\left[ \log \,x\int{x\,dx-\int{\left( \frac{d}{dx}\log \,x\int{x\,dx} \right)dx}} \right]\]

\[=\frac{{{x}^{2}}}{2}{{(\log \,x)}^{2}}-\left[ \log \,x\cdot \frac{{{x}^{2}}}{2}-\int{\frac{1}{x}\cdot \frac{{{x}^{2}}}{2}}dx \right]\]

\[=\frac{{{x}^{2}}}{2}{{(\log \,x)}^{2}}-\frac{{{x}^{2}}}{2}\cdot \log x+\int{\frac{x}{2}}dx\]

\[=\frac{{{x}^{2}}}{2}{{(\log \,x)}^{2}}-\frac{{{x}^{2}}}{2}\log x+\frac{{{x}^{2}}}{4}+C\]