Integral exercise asked by Imad Zak and solved by 3 ways (by parts , comparison and substitation )

Exercise:

Compute, $\int_{1}^{e}{{{\ln }^{3}}x\,dx}$

Solution: Let $u={{\ln }^{3}}x\,\,\And \,\,dv=dx\Leftrightarrow du=\frac{3}{x}{{\ln }^{2}}x\,dx\,\And \,\,v=x$

So $\int_{1}^{e}{{{\ln }^{3}}x\,dx}=\left( x{{\ln }^{3}}x \right)_{1}^{e}-3\int_{1}^{e}{{{\ln }^{2}}x\,dx}$

Let $u={{\ln }^{2}}x\,\,\And \,\,dv=dx\Leftrightarrow du=\frac{2}{x}\ln x\,dx\,\And \,\,v=x$

So $\int_{1}^{e}{{{\ln }^{3}}x\,dx}=\left( x{{\ln }^{3}}x \right)_{1}^{e}-3\left( x{{\ln }^{2}}x \right)_{1}^{e}+6\int_{1}^{e}{\ln x\,dx}$

                 $=\left( x{{\ln }^{3}}x \right)_{1}^{e}-3\left( x{{\ln }^{2}}x \right)_{1}^{e}+6\left( x\ln x \right)_{1}^{e}-\left( 6x \right)_{1}^{e}$

                $=e{{\ln }^{3}}e-3e{{\ln }^{2}}e+6e\ln e-6e+6$

               $=e-3e+6e-6e+6=6-2e$

Or: Let $F\left( x \right)=x\left( a{{\ln }^{3}}x+b{{\ln }^{2}}x+c\ln x+d \right)=\int{{{\ln }^{3}}x\,dx}$

$F'\left( x \right)=a{{\ln }^{3}}x+b{{\ln }^{2}}x+c\ln x+d+\left( \frac{3a}{x}{{\ln }^{2}}x+\frac{2b}{x}\ln x+\frac{c}{x} \right)x$

         $=a{{\ln }^{3}}x+b{{\ln }^{2}}x+c\ln x+d+3a{{\ln }^{2}}x+2b\ln x+c$

         $=a{{\ln }^{3}}x+\left( 3a+b \right){{\ln }^{2}}x+\left( c+2b \right)\ln x+c+d$

So $a=1\,\,,3a+b=0\,,\,\,c+2b=0\,\,,c+d=0$ Thus $\left\{ a,b,c,d \right\}=\left\{ 1,-3,6,-6 \right\}$

Hence $F\left( x \right)=x\left( {{\ln }^{3}}x-3{{\ln }^{2}}x+6\ln x-6 \right)$

Or: we have $\int_{1}^{e}{{{\ln }^{3}}x\,dx}=\int_{1}^{e}{{{\ln }^{3}}x}\,{{e}^{\ln x}}d\left( \ln x \right)=\int_{0}^{1}{{{u}^{3}}{{e}^{u}}du}$

So $\int_{0}^{1}{{{u}^{3}}{{e}^{u}}\,du}=\left( {{u}^{3}}{{e}^{u}} \right)_{0}^{1}-3\int_{0}^{1}{{{u}^{2}}{{e}^{u}}du}=\left( {{u}^{3}}{{e}^{u}} \right)_{0}^{1}-3\left( {{u}^{2}}{{e}^{u}} \right)_{0}^{1}+6\int_{0}^{1}{u{{e}^{u}}du}$

                 \[=\left( {{u}^{3}}{{e}^{u}} \right)_{0}^{1}-3\left( {{u}^{2}}{{e}^{u}} \right)_{0}^{1}+6\left( u{{e}^{u}} \right)_{0}^{1}-6\left( {{e}^{u}} \right)_{0}^{1}=\left( {{e}^{u}}\left( {{u}^{3}}-3{{u}^{2}}+6u-6 \right) \right)_{0}^{1}\]

                $=e\left( 1-3+6-6 \right)-1\left( -6 \right)=6-2e$




*____________________
2nd Method credit to Imad Zak
3rd Method credit to Kunihiko Chikaya