Question Number 27805 by abdo imad last updated on 15/Jan/18
$${find}\:\:\int_{\mathrm{1}} ^{\propto} \:\:\frac{{arctan}\left(\alpha{x}\right)}{{x}^{\mathrm{2}} }\:. \\ $$
Commented by abdo imad last updated on 16/Jan/18
![let put I= ∫_1 ^∝ ((arctan(αx))/x^2 )dx integrate per parts I=[ −(1/x) arctan(αx)]_1 ^(+∝) − ∫_1 ^(+∝) −(1/x) (α/(1+α^2 x^2 ))dx I= artan(α) +α∫_1 ^(+∝) (dx/(x(1+α^2 x^2 ))) we use the ch. x=(1/α)t ∫_1 ^(+∝) (dx/(x(1+α^2 x^2 ))) = ∫_α ^(+∝) (((1/α)dt)/((1/α)t(1+t^2 ))) = ∫_α ^(+∝) (dt/(t(1+t^2 ))) but(1/(t (1+t^2 ))) = (1/t) − (t/(1+t^2 )) ⇒ ∫ (dt/(t(1+t^2 )))= ln/t/ −(1/2)ln(1+t^2 ) +k = ln/(t/(√(1+t^2 )))/ so ∫_α ^(+∝) (dt/(t(1+t^2 ))) =[ ln/ (t/(√(1+t^2 )))/]_α ^(+∝) =−ln/ (α/(√(1+α^(2 ) )))/ I=artan(α)−α ln/ (α/(√(1+α^2 )))/ .](Q27914.png)
$${let}\:{put}\:{I}=\:\int_{\mathrm{1}} ^{\propto} \:\frac{{arctan}\left(\alpha{x}\right)}{{x}^{\mathrm{2}} }{dx}\:\:{integrate}\:{per}\:{parts} \\ $$$${I}=\left[\:−\frac{\mathrm{1}}{{x}}\:{arctan}\left(\alpha{x}\right)\right]_{\mathrm{1}} ^{+\propto} \:−\:\int_{\mathrm{1}} ^{+\propto} −\frac{\mathrm{1}}{{x}}\:\frac{\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} {x}^{\mathrm{2}} }{dx} \\ $$$${I}=\:{artan}\left(\alpha\right)\:+\alpha\int_{\mathrm{1}} ^{+\propto} \:\:\:\frac{{dx}}{{x}\left(\mathrm{1}+\alpha^{\mathrm{2}} {x}^{\mathrm{2}} \right)}\:{we}\:{use}\:{the}\:{ch}.\:{x}=\frac{\mathrm{1}}{\alpha}{t} \\ $$$$\int_{\mathrm{1}} ^{+\propto} \:\:\frac{{dx}}{{x}\left(\mathrm{1}+\alpha^{\mathrm{2}} {x}^{\mathrm{2}} \right)}\:=\:\int_{\alpha} ^{+\propto} \:\:\frac{\frac{\mathrm{1}}{\alpha}{dt}}{\frac{\mathrm{1}}{\alpha}{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:=\:\int_{\alpha} ^{+\propto} \:\:\:\frac{{dt}}{{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)} \\ $$$$\:{but}\frac{\mathrm{1}}{{t}\:\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:=\:\frac{\mathrm{1}}{{t}}\:−\:\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\Rightarrow\:\int\:\:\frac{{dt}}{{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}=\:{ln}/{t}/\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\:+{k} \\ $$$$=\:{ln}/\frac{{t}}{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}/\:\:{so}\:\:\:\int_{\alpha} ^{+\propto} \:\:\frac{{dt}}{{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:=\left[\:{ln}/\:\frac{{t}}{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}/\right]_{\alpha} ^{+\propto} \\ $$$$=−{ln}/\:\frac{\alpha}{\sqrt{\mathrm{1}+\alpha^{\mathrm{2}\:} }}/ \\ $$$${I}={artan}\left(\alpha\right)−\alpha\:{ln}/\:\frac{\alpha}{\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} }}/\:\:.\: \\ $$$$ \\ $$