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Question Number 92138 by hore last updated on 05/May/20

∫(dx/(1+x^2 ))

$$\int\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$

Answered by john santu last updated on 05/May/20

tan^(−1) (x) + c

$$\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)\:+\:\mathrm{c}\: \\ $$

Answered by Rio Michael last updated on 07/May/20

∫ (dx/(1 + x^2 )) let x = tan θ ⇒ (dx/dθ) = sec^2 θ ⇒ ∫(dx/(1+x^2 )) = ∫ (1/(1+ tan^2 θ)) sec^2 θ dθ = ∫((sec^2 θ)/(sec^2 θ))dθ _(1 + tan^2 θ = sec^2 θ) = ∫ dθ = θ + C but x = tan θ ⇒ θ = tan^(−1) x ⇒∫ (dx/(1 + x^2 )) = tan^(−1) x + C

$$\int\:\frac{{dx}}{\mathrm{1}\:+\:{x}^{\mathrm{2}} } \\ $$$$\mathrm{let}\:{x}\:=\:\mathrm{tan}\:\theta\:\Rightarrow\:\frac{{dx}}{{d}\theta}\:=\:\mathrm{sec}^{\mathrm{2}} \theta \\ $$$$\Rightarrow\:\int\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\:\int\:\frac{\mathrm{1}}{\mathrm{1}+\:\mathrm{tan}^{\mathrm{2}} \theta}\:\:\mathrm{sec}^{\mathrm{2}} \theta\:{d}\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\int\frac{\mathrm{sec}^{\mathrm{2}} \theta}{\mathrm{sec}^{\mathrm{2}} \theta}{d}\theta\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:_{\mathrm{1}\:+\:\mathrm{tan}^{\mathrm{2}} \theta\:=\:\mathrm{sec}^{\mathrm{2}} \theta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\int\:{d}\theta\:=\:\theta\:+\:{C} \\ $$$$\mathrm{but}\:{x}\:=\:\mathrm{tan}\:\theta\:\Rightarrow\:\theta\:=\:\mathrm{tan}^{−\mathrm{1}} {x} \\ $$$$\Rightarrow\int\:\frac{{dx}}{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\:=\:\mathrm{tan}^{−\mathrm{1}} {x}\:+\:{C} \\ $$

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