Integral exercise asked by Dan Sitaru in the calculus math group ( Substitution_Long_Division)

Exercise:

Integrate, $\int{\frac{2x+2n+1}{\frac{1}{{{\left( x+n+1 \right)}^{2}}}+\frac{1}{{{\left( x+n \right)}^{2}}}+1}dx}$ ,   $n\in \mathbb{N}$

Solution: Let $w=x+n+1\Leftrightarrow dw=dx\,\,\And \,\,x+n=w-1$

So $\int{\frac{2x+2n+1}{\frac{1}{{{\left( x+n+1 \right)}^{2}}}+\frac{1}{{{\left( x+n \right)}^{2}}}+1}dx}=\int{\frac{2\left( w-1 \right)+1}{\frac{1}{{{w}^{2}}}+\frac{1}{{{\left( w-1 \right)}^{2}}}+1}dw}$

$=\int{\frac{2w-1}{\frac{1}{{{w}^{2}}}+\frac{1}{{{\left( w-1 \right)}^{2}}}+1}dw}=\int{\frac{2w-1}{\frac{{{\left( {{w}^{2}}-w+1 \right)}^{2}}}{{{\left( w-1 \right)}^{2}}{{w}^{2}}}}dw}=\int{\frac{\left( 2w-1 \right)\left( {{w}^{2}}{{\left( w-1 \right)}^{2}} \right)}{{{\left( {{w}^{2}}-w+1 \right)}^{2}}}dw}$

By long – division we get ( since degree of numerator > degree of denominator  )

$\frac{\left( 2w-1 \right)\left( {{w}^{2}}{{\left( w-1 \right)}^{2}} \right)}{{{\left( {{w}^{2}}-w+1 \right)}^{2}}}=2w-1+\frac{2w-1}{{{\left( {{w}^{2}}-w+1 \right)}^{2}}}-\frac{2\left( 2w-1 \right)}{{{w}^{2}}-w+1}$

So$\int{\frac{\left( 2w-1 \right)\left( {{w}^{2}}{{\left( w-1 \right)}^{2}} \right)}{{{\left( {{w}^{2}}-w+1 \right)}^{2}}}dw}=2\int{wdw-\int{dw+\int{\frac{2w-1}{{{\left( {{w}^{2}}-w+1 \right)}^{2}}}dw-2\int{\frac{2w-1}{{{w}^{2}}-w+1}dw}}}}$

                                       $={{w}^{2}}-w+\int{\frac{d\left( {{w}^{2}}-w+1 \right)}{{{\left( {{w}^{2}}-w+1 \right)}^{2}}}-2\int{\frac{d\left( {{w}^{2}}-w+1 \right)}{{{w}^{2}}-w+1}}}$

                                       $={{w}^{2}}-w-\frac{1}{\left( {{w}^{2}}-w+1 \right)}-2\ln \left| {{w}^{2}}-w+1 \right|+c$

Thus $\int{\frac{2x+2n+1}{\frac{1}{{{\left( x+n+1 \right)}^{2}}}+\frac{1}{{{\left( x+n \right)}^{2}}}+1}dx}$

$={{\left( x+n+1 \right)}^{2}}-\left( x+n+1 \right)-\frac{1}{{{\left( x+n+1 \right)}^{2}}-\left( x+n \right)}-2\ln \left| {{\left( x+n+1 \right)}^{2}}-\left( x+n \right) \right|+c$