Question Number 89918 by jagoll last updated on 20/Apr/20
$$\int\:\frac{\mathrm{x}\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} }\:\mathrm{dx}\: \\ $$
Commented by john santu last updated on 20/Apr/20
![[ u = tan^(−1) (x) ⇒du = (dx/(1+x^2 ))] v = ∫ ((x dx)/((1+x^2 )^(3/2) )) = (1/2)∫ ((d(1+x^2 ))/((1+x^2 )^(3/2) )) v = − (1/( (√(1+x^2 )))) ] ⇒ I = −((tan^(−1) (x))/( (√(1+x^2 )))) + ∫ (dx/((1+x^2 )^(3/2) )) let J = ∫ (dx/((1+x^2 )^(3/2) )) x = tan p ⇒dx = sec^2 p dp J = ∫ ((sec^2 p dp)/(sec^3 p)) = ∫ cos p dp J = sin p = (x/( (√(1+x^2 )))) therefore we get ∫ ((x tan^(−1) (x) dx)/((1+x^2 )^(3/2) )) = −((tan^(−1) (x))/( (√(1+x^2 )))) +(x/( (√(1+x^2 )))) + c = ((x−tan^(−1) (x))/( (√(1+x^2 )))) + c](https://www.tinkutara.com/question/Q89921.png)
$$\left[\:{u}\:=\:\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\:\Rightarrow{du}\:=\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\right] \\ $$$${v}\:=\:\int\:\frac{{x}\:{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{{d}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} } \\ $$$$\left.{v}\:=\:−\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:\right]\: \\ $$$$\Rightarrow\:{I}\:=\:−\frac{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:+\:\int\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} } \\ $$$${let}\:{J}\:=\:\int\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} } \\ $$$${x}\:=\:\mathrm{tan}\:{p}\:\Rightarrow{dx}\:=\:\mathrm{sec}\:^{\mathrm{2}} {p}\:{dp} \\ $$$${J}\:=\:\int\:\frac{\mathrm{sec}\:^{\mathrm{2}} {p}\:{dp}}{\mathrm{sec}\:^{\mathrm{3}} {p}}\:=\:\int\:\mathrm{cos}\:{p}\:{dp} \\ $$$${J}\:=\:\mathrm{sin}\:{p}\:=\:\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$${therefore}\:{we}\:{get}\: \\ $$$$\int\:\frac{{x}\:\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\:{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} }\:=\:−\frac{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:+\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:+\:{c}\: \\ $$$$=\:\frac{{x}−\mathrm{tan}^{−\mathrm{1}} \left({x}\right)}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:+\:{c}\: \\ $$