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Question Number 81482 by Bash last updated on 13/Feb/20

$$Evaluate{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dx}{x^2+4x+13}.$$

Commented by abdomathmax last updated on 13/Feb/20

$$I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x^2+4x+13}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x^2+4x+4+9}$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{{(x+2)}^2+9}{=}_{x+2=3u}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{3du}{9(u^2+1)}$$$$=\frac{1}{3}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{du}{u^2+1}=\frac{1}{3}{\left[arctanu\right]}_{-\mathrm{\infty }}^{+\mathrm{\infty }}=\frac{1}{3}(\frac{\pi }{2}-(-\frac{\pi }{2}))$$$$=\frac{\pi }{3}$$

Answered by MJS last updated on 13/Feb/20

$$\int \frac{dx}{x^2+4x+13}=$$$$[t=\frac{x+2}{3}\to dx=3dt]$$$$=\frac{1}{3}\int \frac{dt}{t^2+1}=\frac{1}{3}\arctan t=\frac{1}{3}\arctan \frac{x+2}{3}+C$$$$\underset{-\mathrm{\infty }}{\stackrel{+\mathrm{\infty }}{\int }}\frac{dx}{x^2+4x+13}=\frac{\pi }{3}$$

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