Question-54209

Question Number 54209 by cesar.marval.larez@gmail.com last updated on 31/Jan/19

Answered by Joel578 last updated on 31/Jan/19

I = ∫ ((tan^(−1) x)/(1 + x^2 )) dx u = tan^(−1) x → du = (dx/(1 + x^2 )) dv = (dx/(1 + x^2 )) → v = tan^(−1) x I = (tan^(−1) x)^2 − ∫ ((tan^(−1) x)/(1 + x^2 )) dx = (tan^(−1) x)^2 − I 2I = (tan^(−1) x)^2 I = (1/2)(tan^(−1) x)^2 + C

$${I}\:=\:\int\:\frac{\mathrm{tan}^{−\mathrm{1}} {x}}{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\:{dx} \\ $$$${u}\:=\:\mathrm{tan}^{−\mathrm{1}} \:{x}\:\:\:\rightarrow\:{du}\:=\:\frac{{dx}}{\mathrm{1}\:+\:{x}^{\mathrm{2}} } \\ $$$${dv}\:=\:\frac{{dx}}{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\:\:\:\rightarrow\:\:{v}\:=\:\mathrm{tan}^{−\mathrm{1}} \:{x} \\ $$$${I}\:=\:\left(\mathrm{tan}^{−\mathrm{1}} \:{x}\right)^{\mathrm{2}} \:−\:\int\:\frac{\mathrm{tan}^{−\mathrm{1}} \:{x}}{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\:{dx} \\ $$$$\:\:\:=\:\left(\mathrm{tan}^{−\mathrm{1}} \:{x}\right)^{\mathrm{2}} \:−\:{I} \\ $$$$\mathrm{2}{I}\:=\:\left(\mathrm{tan}^{−\mathrm{1}} \:{x}\right)^{\mathrm{2}} \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{tan}^{−\mathrm{1}} \:{x}\right)^{\mathrm{2}} \:+\:{C} \\ $$

Answered by Prithwish sen last updated on 31/Jan/19

let, tan^(−1) x = t then (dx/(1+x^2 )) = dt ∴ ∫ ((tan^(−1) x)/(1+x^2 ))dx = ∫ t dt = ((t2)/2) + c = (1/2) (tan^(−1) x)^2 + c

$$ \\ $$$$\mathrm{let},\:\mathrm{tan}^{−\mathrm{1}} \mathrm{x}\:=\:\mathrm{t} \\ $$$$\mathrm{then}\:\frac{\mathrm{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\:{dt} \\ $$$$\therefore\:\int\:\frac{\mathrm{tan}^{−\mathrm{1}} \mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:=\:\int\:\mathrm{t}\:\mathrm{dt} \\ $$$$=\:\frac{\mathrm{t2}}{\mathrm{2}}\:+\:\mathrm{c} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left(\mathrm{tan}^{−\mathrm{1}} \mathrm{x}\right)^{\mathrm{2}} \:+\:\mathrm{c} \\ $$

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