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Question Number 109891 by bemath last updated on 26/Aug/20

((Δbe▽)/(math)) ∫ ((arc tan x)/((1+(1/x^2 )))) dx ?

$$\:\:\frac{\Delta{be}\bigtriangledown}{{math}} \\ $$$$\int\:\frac{\mathrm{arc}\:\mathrm{tan}\:{x}}{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}\:{dx}\:? \\ $$

Answered by john santu last updated on 26/Aug/20

((JS)/(★■★)) I=∫ ((x^2 tan^(−1) (x))/(x^2 +1)) dx = ∫ (({(x^2 +1)−1}tan^(−1) (x))/(x^2 +1)) dx = ∫ (((x^2 +1)tan^(−1) (x))/(x^2 +1))dx−∫ ((tan^(−1) (x))/(x^2 +1)) dx =∫ tan^(−1) (x) dx−∫ tan^(−1) (x) d(tan^(−1) (x)) = x.tan^(−1) (x)−∫ (x/(x^2 +1)) dx−(1/2)(tan^(−1) (x))^2 =x.tan^(−1) (x)−(1/2)ln (x^2 +1)−(1/2)(tan^(−1) (x))^2 + c

$$\:\:\frac{{JS}}{\bigstar\blacksquare\bigstar} \\ $$$${I}=\int\:\frac{{x}^{\mathrm{2}} \:\mathrm{tan}^{−\mathrm{1}} \left({x}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx}\: \\ $$$$=\:\int\:\frac{\left\{\left({x}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{1}\right\}\mathrm{tan}^{−\mathrm{1}} \left({x}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx} \\ $$$$=\:\int\:\frac{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{tan}^{−\mathrm{1}} \left({x}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}−\int\:\frac{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx} \\ $$$$=\int\:\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\:{dx}−\int\:\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\:{d}\left(\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\right) \\ $$$$=\:{x}.\mathrm{tan}^{−\mathrm{1}} \left({x}\right)−\int\:\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\right)^{\mathrm{2}} \\ $$$$={x}.\mathrm{tan}^{−\mathrm{1}} \left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\right)^{\mathrm{2}} +\:{c}\: \\ $$

Commented by bemath last updated on 26/Aug/20

a farmer math

Commented by Rasheed.Sindhi last updated on 26/Aug/20

Do you see any hoes there? :)

$$\left.\mathcal{D}{o}\:{you}\:{see}\:{any}\:{hoes}\:{there}?\::\right) \\ $$

Commented by bemath last updated on 26/Aug/20

no sir. hahahaha

$${no}\:{sir}.\:{hahahaha} \\ $$

Commented by Rasheed.Sindhi last updated on 26/Aug/20

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