Integral Exercise using by parts tech asked by Katrina Linda

Exercise:

Integrate, $\int{\frac{{{\left( \ln x \right)}^{2}}}{{{x}^{2}}}dx}$

Solution: Let $u=\ln x\Rightarrow du=\frac{1}{x}dx$ and \({{e}^{u}}=x\Leftrightarrow {{e}^{{{u}^{2}}}}={{x}^{2}}\)

So $\int{\frac{{{\left( \ln x \right)}^{2}}}{{{x}^{2}}}dx}=\int{\frac{{{u}^{2}}x}{{{x}^{2}}}du}=\int{\frac{{{u}^{2}}}{{{e}^{u}}}du=\int{{{u}^{2}}{{e}^{-u}}du}}$

Let $U={{u}^{2}}\,\,\And \,\,dV={{e}^{-u}}du$ $\Leftrightarrow dU=2u\,du\,\,\And \,V=-{{e}^{-u}}$

So $\int{{{u}^{2}}{{e}^{-u}}du}=-{{u}^{2}}{{e}^{-u}}+2\int{u{{e}^{-u}}}du$

Again Take $F=u\,\,\And \,\,dG={{e}^{-u}}du$ $\Leftrightarrow dF=du\,\,\And \,\,G=-{{e}^{-u}}$

So $\int{u{{e}^{-u}}du}=-u{{e}^{-u}}+\int{{{e}^{-u}}du}=-u{{e}^{-u}}-{{e}^{-u}}+c$

i.e $\int{{{u}^{2}}{{e}^{-u}}du}=-{{u}^{2}}{{e}^{-u}}-2u{{e}^{-u}}-2{{e}^{-u}}+c$

thus $\int{\frac{{{\left( \ln x \right)}^{2}}}{{{x}^{2}}}dx}=-{{\left( \ln x \right)}^{2}}{{e}^{\ln {{x}^{-1}}}}-2\left( \ln x \right){{e}^{\ln {{x}^{-1}}}}-2{{e}^{\ln {{x}^{-1}}}}+c$

                                 $=-\frac{{{\left( \ln x \right)}^{2}}}{x}-\frac{2\left( \ln x \right)}{x}-\frac{2}{x}+c$