Question

Integrate the function: $\frac{\sqrt{x^{2}+1}[log(x^{2}+1)-2logx]}{x^{4}}$

Answer 1

\frac{\sqrt{x^{2}+1}[log(x^{2}+1)-2logx]}{x^{4}}$$=\frac{\sqrt{x^{2}+1}}{x4}[log(x^{2}+1)-logx^{2}]

$=\frac{\sqrt{x^{2}+1}}{x^{4}}[log(\frac{x^{2}+1}{x^{2}})]$

$=\sqrt{\frac{x^{2}+1}{x^{4}}}log(1+\frac{1}{x^{2}})$

$=\frac{1}{x^{3}}\sqrt{\frac{x^{2}+1}{x^{2}}}log(1+\frac{1}{x^{2}})$

$=\frac{1}{x^{3}}\sqrt{x^{2}+\frac{1}{x^{2}}}log(1+\frac{1}{x^{2}})$

$Let 1+\frac{1}{x^{2}}=t⇒\frac{-2}{x^{3}}dx=dt$

$∴I=∫\frac{1}{x^{3}}\sqrt{1+\frac{1}{{x^{2}log}}(1+\frac{1}{x^{2}}})dx$

$=\frac{-1}{2}∫\sqrt{t}\space logt\space dt$

$=\frac{-1}{2}∫t^{\frac{1}{2}}.logt\space dt$

Integrating by parts,we obtain

$I=\frac{-1}{2}[logt.∫t^{\frac{1}{2}}dt-{(\frac{d}{dt}logt)∫t^{\frac{1}{2}}dt}\space dt]$

$=\frac{-1}{2}[logt.\frac{t^{\frac{3}{2}}}{\frac{3}{2}}-∫\frac{1}{t}.\frac{t^{\frac{3}{2}}}{\frac{3}{2}}dt]$

$=\frac{-1}{2}[\frac{2}{3}t^{\frac{3}{2}}logt-\frac{2}{3}∫t^{\frac{1}{2}}dt]$

$=-\frac{1}{2}[\frac{2}{3}t^{\frac{3}{2}}logt-\frac{4}{9}t^{\frac{3}{2}}]$

$=-\frac{1}{3}t^{\frac{3}{2}}logt+\frac{2}{9}t^{\frac{3}{2}}$

$=-\frac{1}{3}t^{\frac{3}{2}}[logt-\frac{2}{3}]$

$=-\frac{1}{3}(1+\frac{1}{x^{2}})^{\frac{3}{2}}[log(1+\frac{1}{x^{2}})-\frac{2}{3}]+C$