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Question Number 99044 by mashallah last updated on 18/Jun/20

∫(1/(x^2 +1))dx=?

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

Answered by bemath last updated on 18/Jun/20

arc tan x + c

$$\mathrm{arc}\:\mathrm{tan}\:\mathrm{x}\:+\:\mathrm{c} \\ $$

Answered by Mr.D.N. last updated on 18/Jun/20

  I=∫(1/(x^2 +1))dx    Put, x=tan θ      ⇒ θ = tan^(−1) x       dx=sec^2 θdθ     I= ∫ (1/(tan^2 θ+1)).sec^2 θdθ      = ∫ (1/(sec^2 θ))sec^2 θdθ = ∫dθ    =  θ +C    = tan^(−1) x+C.

$$\:\:\mathrm{I}=\int\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx} \\ $$$$\:\:\mathrm{Put},\:\mathrm{x}=\mathrm{tan}\:\theta\:\:\:\:\:\:\Rightarrow\:\theta\:=\:\mathrm{tan}^{−\mathrm{1}} \mathrm{x} \\ $$$$\:\:\:\:\:\mathrm{dx}=\mathrm{sec}^{\mathrm{2}} \theta\mathrm{d}\theta \\ $$$$\:\:\:\mathrm{I}=\:\int\:\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \theta+\mathrm{1}}.\mathrm{sec}^{\mathrm{2}} \theta\mathrm{d}\theta \\ $$$$\:\:\:\:=\:\int\:\frac{\mathrm{1}}{\mathrm{sec}^{\mathrm{2}} \theta}\mathrm{sec}^{\mathrm{2}} \theta\mathrm{d}\theta\:=\:\int\mathrm{d}\theta \\ $$$$\:\:=\:\:\theta\:+\mathrm{C} \\ $$$$\:\:=\:\mathrm{tan}^{−\mathrm{1}} \mathrm{x}+\mathrm{C}. \\ $$

Answered by abdomathmax last updated on 18/Jun/20

complex method   I =∫ (dx/((x−i)(x+i))) =(1/(2i))∫  ((1/(x−i))−(1/(x+i)))dx  =(1/(2i))ln(((x−i)/(x+i))) +C

$$\mathrm{complex}\:\mathrm{method}\: \\ $$$$\mathrm{I}\:=\int\:\frac{\mathrm{dx}}{\left(\mathrm{x}−\mathrm{i}\right)\left(\mathrm{x}+\mathrm{i}\right)}\:=\frac{\mathrm{1}}{\mathrm{2i}}\int\:\:\left(\frac{\mathrm{1}}{\mathrm{x}−\mathrm{i}}−\frac{\mathrm{1}}{\mathrm{x}+\mathrm{i}}\right)\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2i}}\mathrm{ln}\left(\frac{\mathrm{x}−\mathrm{i}}{\mathrm{x}+\mathrm{i}}\right)\:+\mathrm{C} \\ $$

Commented by abdomathmax last updated on 18/Jun/20

arctanx =(1/(2i))ln(((1+ix)/(1−ix)))

$$\mathrm{arctanx}\:=\frac{\mathrm{1}}{\mathrm{2i}}\mathrm{ln}\left(\frac{\mathrm{1}+\mathrm{ix}}{\mathrm{1}−\mathrm{ix}}\right) \\ $$

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