Question Number 14284 by tawa tawa last updated on 30/May/17
Answered by Tinkutara last updated on 30/May/17
![∫_((−3)/2) ^∞ (dx/(4x^2 + 12x + 13)) = (1/4)∫_((−3)/2) ^∞ (dx/(x^2 + 3x + ((13)/4))) = (1/4)∫_((−3)/2) ^∞ (dx/((x + (3/2))^2 + 1^2 )) = (1/4)[tan^(−1) (x + (3/2))]_((−3)/2) ^∞ = (1/4)[tan^(−1) ∞ − tan^(−1) 0] = (1/4)[(π/2)] = (π/8)](https://www.tinkutara.com/question/Q14285.png)
$$\int_{\frac{−\mathrm{3}}{\mathrm{2}}} ^{\infty} \frac{{dx}}{\mathrm{4}{x}^{\mathrm{2}} \:+\:\mathrm{12}{x}\:+\:\mathrm{13}}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\int_{\frac{−\mathrm{3}}{\mathrm{2}}} ^{\infty} \frac{{dx}}{{x}^{\mathrm{2}} \:+\:\mathrm{3}{x}\:+\:\frac{\mathrm{13}}{\mathrm{4}}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\int_{\frac{−\mathrm{3}}{\mathrm{2}}} ^{\infty} \frac{{dx}}{\left({x}\:+\:\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\:\mathrm{1}^{\mathrm{2}} } \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{tan}^{−\mathrm{1}} \:\left({x}\:+\:\frac{\mathrm{3}}{\mathrm{2}}\right)\right]_{\frac{−\mathrm{3}}{\mathrm{2}}} ^{\infty} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{tan}^{−\mathrm{1}} \infty\:−\:\mathrm{tan}^{−\mathrm{1}} \:\mathrm{0}\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{\pi}{\mathrm{2}}\right]\:=\:\frac{\pi}{\mathrm{8}} \\ $$
Commented by tawa tawa last updated on 30/May/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by tawa tawa last updated on 30/May/17
$$\mathrm{please}\:\mathrm{check}\:\mathrm{the}\:\mathrm{line}\:\mathrm{3}.\:\mathrm{is}\:\mathrm{it}\:\mathrm{correct}\:?\:\mathrm{or}\:\mathrm{mistake} \\ $$
Commented by tawa tawa last updated on 30/May/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$