Integral exercise asked in many math groups by Dan Situra

Exercise:

Compute, $\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\int_{1}^{n}{\frac{{{x}^{4}}+4{{x}^{3}}+12{{x}^{2}}+9x}{{{\left( x+3 \right)}^{5}}-{{x}^{5}}-243}dx}$

Solution: we have ${{x}^{4}}+4{{x}^{3}}+12{{x}^{2}}+9x=x\left( {{x}^{3}}+4{{x}^{2}}+12x+9 \right)$

Observe that $x=-1$ is a root so ${{x}^{3}}+4{{x}^{2}}+12x+9=\left( x+1 \right)\left( {{x}^{2}}+3x+9 \right)$

Thus ${{x}^{4}}+4{{x}^{3}}+12{{x}^{2}}+9x=x\left( x+1 \right)\left( {{x}^{2}}+3x+9 \right)$

Observe that ${{\left( x+3 \right)}^{5}}-{{x}^{5}}-243={{\left( x+3 \right)}^{5}}-{{x}^{5}}-{{3}^{5}}={{\left( x+3 \right)}^{5}}-\left( {{x}^{5}}+{{3}^{5}} \right)$

By using, ${{x}^{n}}+{{y}^{n}}=\left( x+y \right)\left( {{x}^{n-1}}-{{x}^{n-2}}y+...-x{{y}^{n-2}}+{{y}^{n-1}} \right)\,\,\,,\,\,n\,\,odd$

So ${{x}^{5}}+{{3}^{5}}=\left( x+3 \right)\left( {{x}^{4}}-3{{x}^{3}}+{{3}^{2}}{{x}^{2}}-{{3}^{3}}x+{{3}^{4}} \right)$

Hence ${{\left( x+3 \right)}^{5}}-\left( {{x}^{5}}+{{3}^{5}} \right)={{\left( x+3 \right)}^{5}}-\left( x+3 \right)\left( {{x}^{4}}-3{{x}^{3}}+9{{x}^{2}}-27x+81 \right)$

$=\left( x+3 \right)\left( {{\left( x+3 \right)}^{4}}-{{x}^{4}}+3{{x}^{3}}-9{{x}^{2}}+27x-81 \right)=15x\left( x+3 \right)\left( {{x}^{2}}+3x+9 \right)$

So $\frac{{{x}^{4}}+4{{x}^{3}}+12{{x}^{2}}+9x}{{{\left( x+3 \right)}^{5}}-{{x}^{5}}-243}=\frac{x\left( x+1 \right)\left( {{x}^{2}}+3x+9 \right)}{15x\left( x+3 \right)\left( {{x}^{2}}+3x+9 \right)}=\frac{x+1}{15\left( x+3 \right)}$

So $\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\int_{1}^{n}{\frac{{{x}^{4}}+4{{x}^{3}}+12{{x}^{2}}+9x}{{{\left( x+3 \right)}^{5}}-{{x}^{5}}-243}dx}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\int_{1}^{n}{\frac{x+1}{15\left( x+3 \right)}dx}$

But $\int{\frac{x+1}{x+3}dx}=\int{\frac{x+1-3+3}{x+3}dx}=\int{\frac{x+3}{x+3}dx-2\int{\frac{dx}{x+3}=x-2\ln \left| x+3 \right|+c}}$

Hence $\int_{1}^{n}{\frac{x+1}{15\left( x+3 \right)}dx}=\frac{1}{15}\left( x-2\ln \left( x+3 \right) \right)_{1}^{n}=\frac{1}{15}\left( n-2\ln \left( n+3 \right)-\left( 1-2\ln 4 \right) \right)$

Thus $\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\int_{1}^{n}{\frac{x+1}{15\left( x+3 \right)}dx}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{15}\left( 1-\frac{2\ln \left( n+3 \right)}{n}-\frac{1-2\ln 4}{n} \right)=\frac{1}{15}$