Question Number 19452 by icyfalcon999 last updated on 11/Aug/17
Answered by Tinkutara last updated on 11/Aug/17
![∫(dx/(x^2 + 1)) = tan^(−1) x [tan^(−1) x]_(−∞) ^∞ = (π/2) − (−(π/2)) = π](https://www.tinkutara.com/question/Q19454.png)
$$\int\frac{{dx}}{{x}^{\mathrm{2}} \:+\:\mathrm{1}}\:=\:\mathrm{tan}^{−\mathrm{1}} {x} \\ $$$$\left[\mathrm{tan}^{−\mathrm{1}} {x}\right]_{−\infty} ^{\infty} \:=\:\frac{\pi}{\mathrm{2}}\:−\:\left(−\frac{\pi}{\mathrm{2}}\right)\:=\:\pi \\ $$
Commented by icyfalcon999 last updated on 11/Aug/17
$$\mathrm{I}\:\mathrm{didn}'\mathrm{t}\:\mathrm{understand}\:,\mathrm{could}\:\mathrm{you}\:\mathrm{include}\:\mathrm{more}\:\mathrm{details}? \\ $$