integral exercise asked in mathematics teacher group by وليد غني

Exercise:

Integrate, $\int{\frac{{{e}^{{{x}^{2}}+4\ln x}}-{{e}^{{{x}^{2}}+3\ln x}}}{x-1}dx}$

Solution: we have ${{e}^{{{x}^{2}}+4\ln x}}={{e}^{{{x}^{2}}}}{{e}^{4\ln x}}={{e}^{{{x}^{2}}}}{{e}^{\ln {{x}^{4}}}}={{x}^{4}}{{e}^{{{x}^{2}}}}$

Similarly for ${{e}^{{{x}^{2}}+3\ln x}}={{e}^{{{x}^{2}}}}{{e}^{\ln {{x}^{3}}}}={{x}^{3}}{{e}^{{{x}^{2}}}}$

Hence ${{e}^{{{x}^{2}}+4\ln x}}-{{e}^{{{x}^{2}}+3\ln x}}={{x}^{4}}{{e}^{{{x}^{2}}}}-{{x}^{3}}{{e}^{{{x}^{2}}}}={{x}^{3}}{{e}^{{{x}^{2}}}}\left( x-1 \right)$

So $\int{\frac{{{e}^{{{x}^{2}}+4\ln x}}-{{e}^{{{x}^{2}}+3\ln x}}}{x-1}dx}=\int{\frac{{{x}^{3}}{{e}^{{{x}^{2}}}}\left( x-1 \right)}{x-1}dx}=\int{{{x}^{3}}{{e}^{{{x}^{2}}}}dx}$

Let $u={{x}^{2}}\Leftrightarrow du=2xdx\Leftrightarrow xdx=\frac{du}{2}$

So $\int{{{x}^{2}}x{{e}^{{{x}^{2}}}}dx}=\int{u{{e}^{u}}\frac{du}{2}}=\frac{1}{2}\int{u{{e}^{u}}du}$

Take $U=u\,\,\And \,\,dv={{e}^{u}}du\Leftrightarrow dU=du\,\,\And \,v={{e}^{u}}$

So $\int{u{{e}^{u}}du}=u{{e}^{u}}-\int{{{e}^{u}}du}=u{{e}^{u}}-{{e}^{u}}={{e}^{u}}\left( u-1 \right)+c$

Thus $\int{{{x}^{3}}{{e}^{{{x}^{2}}}}dx}=\frac{1}{2}{{e}^{{{x}^{2}}}}\left( {{x}^{2}}-1 \right)+c$