2
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Show that

$$\int_{-\infty}^{\infty}\frac{x^2dx}{(x^2+1)^2(x^2+2x+2)}=\frac{7\pi}{50} $$

So I figured since it's an improper integral I should change the limits

$$\lim_{m_1\to-\infty}\int_{m_1}^{0}\frac{x^2dx}{(x^2+1)^2(x^2+2x+2)}+ \lim_{m_2\to\infty}\int_{0}^{m_2}\frac{x^2dx}{(x^2+1)^2(x^2+2x+2)}$$

I'm however not sure how to evaluate this. Any help would be great - thanks.

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  • 2
    $\begingroup$ Are you familiar with partial fractions? $\endgroup$ – David H Jun 12 '14 at 17:49
1
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Hint

Use partial fractions to write out the integral in the form: $$\frac{A x+B}{x^2+1}+\frac{C x+D}{\left(x^2+1\right)^2}+\frac{E+F x}{x^2+2 x+2}$$ You'll find that $a=-6/25, c = 2/5$, etc. Now look at at what remains, and use the fact that: $$\int \frac{1}{x^2+1}dx=\arctan x$$ And substitute $x\rightarrow x+1$ in the last integral to simplify it.

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  • $\begingroup$ You used B twice $\endgroup$ – Ali Caglayan Jun 12 '14 at 19:05
  • $\begingroup$ @Alizter - thanks, fixed. $\endgroup$ – nbubis Jun 12 '14 at 19:28

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