Question Number 199368 by sonukgindia last updated on 02/Nov/23
Answered by qaz last updated on 02/Nov/23
![∫_(−∞) ^(+∞) (x^2 /((x^2 +1)(x^2 +4)))dx =2πi([(z−i)^(−1) ]+[(z−2i)^(−1) ])(z^2 /((z^2 +1)(z^2 +4))) =2πi[z^0 ]((((z+i)^2 )/((z+2i)((z+i)^2 +4)))+(((z+2i)^2 )/(((z+2i)^2 +1)(z+4i)))) =(π/3)](https://www.tinkutara.com/question/Q199373.png)
$$\int_{−\infty} ^{+\infty} \frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{dx} \\ $$$$=\mathrm{2}\pi{i}\left(\left[\left({z}−{i}\right)^{−\mathrm{1}} \right]+\left[\left({z}−\mathrm{2}{i}\right)^{−\mathrm{1}} \right]\right)\frac{{z}^{\mathrm{2}} }{\left({z}^{\mathrm{2}} +\mathrm{1}\right)\left({z}^{\mathrm{2}} +\mathrm{4}\right)} \\ $$$$=\mathrm{2}\pi{i}\left[{z}^{\mathrm{0}} \right]\left(\frac{\left({z}+{i}\right)^{\mathrm{2}} }{\left({z}+\mathrm{2}{i}\right)\left(\left({z}+{i}\right)^{\mathrm{2}} +\mathrm{4}\right)}+\frac{\left({z}+\mathrm{2}{i}\right)^{\mathrm{2}} }{\left(\left({z}+\mathrm{2}{i}\right)^{\mathrm{2}} +\mathrm{1}\right)\left({z}+\mathrm{4}{i}\right)}\right) \\ $$$$=\frac{\pi}{\mathrm{3}} \\ $$
Answered by mr W last updated on 02/Nov/23
![(x^2 /((x^2 +1)(x^2 +4)))=(A/(x^2 +1))+(B/(x^2 +4))=(((A+B)x^2 +4A+B)/((x^2 +1)(x^2 +4))) A+B=1 4A+B=0 ⇒A=−(1/3), B=(4/3) ∫_(−∞) ^(+∞) (x^2 /((x^2 +1)(x^2 +4)))dx =2∫_0 ^(+∞) (x^2 /((x^2 +1)(x^2 +4)))dx =2{−(1/3)∫_0 ^(+∞) (dx/(x^2 +1))+(4/3)∫_0 ^(+∞) (dx/(x^2 +4))} =2{−(1/3)[tan^(−1) x]_0 ^∞ +(4/3)[(1/2)tan^(−1) (x/2)]^∞ _0 } =2{−(1/3)×(π/2)+(4/3)×(1/2)×(π/2)} =(π/3) ✓](https://www.tinkutara.com/question/Q199372.png)
$$\frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}=\frac{{A}}{{x}^{\mathrm{2}} +\mathrm{1}}+\frac{{B}}{{x}^{\mathrm{2}} +\mathrm{4}}=\frac{\left({A}+{B}\right){x}^{\mathrm{2}} +\mathrm{4}{A}+{B}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)} \\ $$$${A}+{B}=\mathrm{1} \\ $$$$\mathrm{4}{A}+{B}=\mathrm{0} \\ $$$$\Rightarrow{A}=−\frac{\mathrm{1}}{\mathrm{3}},\:{B}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\int_{−\infty} ^{+\infty} \frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{+\infty} \frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{dx} \\ $$$$=\mathrm{2}\left\{−\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{+\infty} \frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{4}}{\mathrm{3}}\int_{\mathrm{0}} ^{+\infty} \frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{4}}\right\} \\ $$$$=\mathrm{2}\left\{−\frac{\mathrm{1}}{\mathrm{3}}\left[\mathrm{tan}^{−\mathrm{1}} {x}\right]_{\mathrm{0}} ^{\infty} +\frac{\mathrm{4}}{\mathrm{3}}\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \frac{{x}}{\mathrm{2}}\underset{\mathrm{0}} {\right]}^{\infty} \right\} \\ $$$$=\mathrm{2}\left\{−\frac{\mathrm{1}}{\mathrm{3}}×\frac{\pi}{\mathrm{2}}+\frac{\mathrm{4}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{2}}×\frac{\pi}{\mathrm{2}}\right\} \\ $$$$=\frac{\pi}{\mathrm{3}}\:\checkmark \\ $$