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Question Number 199368 by sonukgindia last updated on 02/Nov/23

Answered by qaz last updated on 02/Nov/23

$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^2}{(x^2+1)(x^2+4)}dx$$$$=2\pi i(\left[{(z-i)}^{-1}\right]+\left[{(z-2i)}^{-1}\right])\frac{z^2}{(z^2+1)(z^2+4)}$$$$=2\pi i\left[z^0\right](\frac{{(z+i)}^2}{(z+2i)({(z+i)}^2+4)}+\frac{{(z+2i)}^2}{({(z+2i)}^2+1)(z+4i)})$$$$=\frac{\pi }{3}$$

Answered by mr W last updated on 02/Nov/23

$$\frac{x^2}{(x^2+1)(x^2+4)}=\frac{A}{x^2+1}+\frac{B}{x^2+4}=\frac{(A+B)x^2+4A+B}{(x^2+1)(x^2+4)}$$$$A+B=1$$$$4A+B=0$$$$\Rightarrow A=-\frac{1}{3},B=\frac{4}{3}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^2}{(x^2+1)(x^2+4)}dx$$$$=2{\int }_0^{+\mathrm{\infty }}\frac{x^2}{(x^2+1)(x^2+4)}dx$$$$=2\{-\frac{1}{3}{\int }_0^{+\mathrm{\infty }}\frac{dx}{x^2+1}+\frac{4}{3}{\int }_0^{+\mathrm{\infty }}\frac{dx}{x^2+4}\}$$$$\text{Missing left or extra right}$$$$=2\{-\frac{1}{3}\times \frac{\pi }{2}+\frac{4}{3}\times \frac{1}{2}\times \frac{\pi }{2}\}$$$$=\frac{\pi }{3}✓$$

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