Definite integral exercise

Exercise:

Show that , $\int_{-\infty }^{\infty }{\frac{x}{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+2x+2 \right)}dx=-\frac{\pi }{5}}$

Solution: we have $\frac{x}{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+2x+2 \right)}=\frac{Ax+B}{{{x}^{2}}+1}+\frac{Cx+D}{{{x}^{2}}+2x+2}$ $=\frac{\left( Ax+B \right)\left( {{x}^{2}}+2x+2 \right)+\left( Cx+D \right)\left( {{x}^{2}}+1 \right)}{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+2x+2 \right)}$

$=\frac{A{{x}^{3}}+2A{{x}^{2}}+2Ax+B{{x}^{2}}+2Bx+2B+C{{x}^{3}}+Cx+D{{x}^{2}}+D}{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+2x+2 \right)}$

$=\frac{{{x}^{3}}\left( A+C \right)+{{x}^{2}}\left( 2A+B+D \right)+x\left( 2A+2B+C \right)+\left( 2B+D \right)}{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+2x+2 \right)}$

After solving we get $A=\frac{1}{5},B=\frac{2}{5},C=\frac{-1}{5},D=\frac{-4}{5}$

$=\frac{\frac{x+2}{5}}{{{x}^{2}}+1}+\frac{\frac{-x-4}{5}}{{{x}^{2}}+2x+1}=\frac{x+2}{5\left( {{x}^{2}}+1 \right)}+\frac{-x-4}{5\left( {{x}^{2}}+2x+2 \right)}$

So $\int_{{}}^{{}}{\frac{x}{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+2x+2 \right)}dx}=\frac{1}{5}\int_{{}}^{{}}{\frac{x+2}{{{x}^{2}}+1}dx}+\frac{1}{5}\int_{{}}^{{}}{\frac{-x-4}{{{x}^{2}}+2x+2}dx}$

Let’s compute the indefinite integrals and in the final we will take a limit

It clear that $\int{\frac{x+2}{{{x}^{2}}+1}dx}=\int{\frac{x}{{{x}^{2}}+1}dx}+\int{\frac{2}{{{x}^{2}}+1}dx}=\int{\frac{x+2}{{{x}^{2}}+1}dx}+2\int{\frac{1}{{{x}^{2}}+1}}$

$=\frac{1}{2}\int{\frac{d\left( {{x}^{2}}+1 \right)}{{{x}^{2}}+1}+2\arctan \left( x \right)+c=\frac{1}{2}\ln \left( {{x}^{2}}+1 \right)+2\arctan \left( x \right)}+c$

Thus $\frac{1}{5}\int{\frac{x+2}{{{x}^{2}}+1}dx}=\frac{1}{10}\ln \left( {{x}^{2}}+1 \right)+\frac{2}{5}\arctan x+c$

Next we need to compute the last term as indefinite integral

We have $\int{\frac{-x-4}{{{x}^{2}}+2x+2}dx}=-\int{\frac{x}{{{x}^{2}}+2x+2}dx-4\int{\frac{dx}{{{x}^{2}}+2x+2}}}$

But ${{x}^{2}}+2x+2={{x}^{2}}+2x+1-1+2={{\left( x+1 \right)}^{2}}+1$

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

Take $u=x+1\Leftrightarrow du=dx\,\And \,\,x=u-1$

So $\int{\frac{x}{{{\left( x+1 \right)}^{2}}+1}dx=\int{\frac{u-1}{{{u}^{2}}+1}du}=\int{\frac{u}{{{u}^{2}}+1}du}-\int{\frac{du}{{{u}^{2}}+1}}}$

$=\frac{1}{2}\int{\frac{d\left( {{u}^{2}}+1 \right)}{{{u}^{2}}+1}-\arctan u+c}=\frac{1}{2}\ln \left( {{u}^{2}}+1 \right)-\arctan u+c$

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

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

And $\int{\frac{4}{{{x}^{2}}+2x+2}dx}=4\int{\frac{dx}{{{\left( x+1 \right)}^{2}}+1}}$

Take $u=x+1\Leftrightarrow du=dx$

Thus $4\int{\frac{dx}{{{\left( x+1 \right)}^{2}}+1}=4\int{\frac{du}{{{u}^{2}}+1}=4\arctan u+c=4\arctan \left( x+1 \right)+c}}$

Hence

$\int{\frac{x}{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+2x+2 \right)}dx}=\frac{1}{10}\ln \left( {{x}^{2}}+1 \right)+\frac{2}{5}\arctan x-\frac{1}{10}\ln \left( {{x}^{2}}+2x+2 \right)-\frac{3}{5}\arctan \left( x+1 \right)+c$

It’s Clear that $\frac{1}{10}\ln \left( {{x}^{2}}+1 \right)-\frac{1}{10}\ln \left( {{x}^{2}}+2x+2 \right)=\frac{1}{10}\ln \left( \frac{{{x}^{2}}+1}{{{x}^{2}}+2x+2} \right)$

Let $F\left( x \right)=\frac{1}{10}\ln \left( \frac{{{x}^{2}}+1}{{{x}^{2}}+2x+2} \right)+\frac{2}{5}\arctan x-\frac{3}{5}\arctan \left( x+1 \right)+c$

Now we need to compute the limits at $-\infty \,\,\And \,\,+\infty $

As we know that $\underset{x\to \infty }{\mathop{\lim }}\,\arctan x=\frac{\pi }{2}\,\,\And \underset{x\to \infty }{\mathop{\lim }}\,\arctan \left( x+1 \right)=\frac{\pi }{2}$  also it is $\frac{-\pi }{2}$ as $x$ tends to $-\infty $

So $\underset{x\to \pm \infty }{\mathop{\lim }}\,\frac{1}{10}\ln \left( \frac{{{x}^{2}}+1}{{{x}^{2}}+2x+1} \right)=\frac{1}{10}\ln \left( 1 \right)=0$ Thus $\underset{x\to \infty }{\mathop{\lim }}\,F\left( x \right)=\frac{-\pi }{10}$ and $\underset{x\to -\infty }{\mathop{\lim }}\,F\left( x \right)=\frac{\pi }{10}$

Thus $\int_{-\infty }^{\infty }{\frac{x}{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+2x+2 \right)}dx}=\underset{x\to \infty }{\mathop{\lim }}\,F\left( x \right)-\underset{x\to -\infty }{\mathop{\lim }}\,F\left( x \right)=\frac{-\pi }{10}-\frac{\pi }{10}=\frac{-\pi }{5}$