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Question-19452




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
$$\int\frac{{dx}}{{x}^{\mathrm{2}} \:+\:\mathrm{1}}\:=\:\mathrm{tan}^{−\mathrm{1}} {x} \\ $$$$\left[\mathrm{tan}^{−\mathrm{1}} {x}\right]_{−\infty} ^{\infty} \:=\:\frac{\pi}{\mathrm{2}}\:−\:\left(−\frac{\pi}{\mathrm{2}}\right)\:=\:\pi \\ $$
Commented by icyfalcon999 last updated on 11/Aug/17
I didn′t understand ,could you include more details?
$$\mathrm{I}\:\mathrm{didn}'\mathrm{t}\:\mathrm{understand}\:,\mathrm{could}\:\mathrm{you}\:\mathrm{include}\:\mathrm{more}\:\mathrm{details}? \\ $$

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