Question Number 179655 by Acem last updated on 31/Oct/22
$${Evaluate}\:\int\mathrm{6}\:\mathrm{arctan}\:\frac{\mathrm{8}}{{w}}\:{dw} \\ $$
Answered by CElcedricjunior last updated on 05/Nov/22
$$\int\mathrm{6}\boldsymbol{{arctan}}\left(\frac{\mathrm{8}}{\boldsymbol{\omega}}\right)\boldsymbol{{d}\omega}=\boldsymbol{{k}} \\ $$$$\boldsymbol{{posons}}\:\begin{cases}{\boldsymbol{{u}}=\boldsymbol{{arctan}}\left(\frac{\mathrm{8}}{\boldsymbol{\omega}}\right)}\\{\boldsymbol{{v}}'=\mathrm{1}}\end{cases}=>\begin{cases}{\boldsymbol{{u}}'=β\frac{\mathrm{8}}{\boldsymbol{\omega}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\mathrm{1}+\left(\frac{\mathrm{8}}{\boldsymbol{\omega}}\right)^{\mathrm{2}} }\:\right)}\\{\boldsymbol{{v}}=\boldsymbol{\omega}}\end{cases} \\ $$$$\boldsymbol{{k}}=\mathrm{6}\boldsymbol{\omega{arctan}}\left(\frac{\mathrm{8}}{\boldsymbol{\omega}}\right)+\mathrm{48}\int\frac{\boldsymbol{\omega}}{\boldsymbol{\omega}^{\mathrm{2}} +\mathrm{64}}\boldsymbol{{d}\omega} \\ $$$$\boldsymbol{{k}}=\mathrm{6}\boldsymbol{{arctan}}\left(\frac{\mathrm{8}}{\boldsymbol{\omega}}\right)+\mathrm{96}\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\omega}^{\mathrm{2}} +\mathrm{64}\right)+\boldsymbol{{cste}} \\ $$$$\int\mathrm{6}\boldsymbol{{arctan}}\left(\frac{\mathrm{8}}{\boldsymbol{\omega}}\right)\boldsymbol{\mathrm{d}\omega}=\mathrm{6}\boldsymbol{{arctan}}\left(\frac{\mathrm{8}}{\boldsymbol{\omega}}\right)+\mathrm{96}\boldsymbol{{ln}}\left(\boldsymbol{\omega}^{\mathrm{2}} +\mathrm{64}\right)+\boldsymbol{{cste}} \\ $$$$……\sqrt{{le}\:{celebre}\:{cedric}\:{junior}……..} \\ $$
Answered by MJS_new last updated on 01/Nov/22
![bβ«arctan (a/x) dx= [((uβ²=1 β u=x)),((v=arctan (x/a) β vβ²=β(a/(x^2 +a^2 )))) ] =bxarctan (x/a) +abβ«(x/(x^2 +a^2 ))dx= =bxarctan (x/a) +((ab)/2)ln (x^2 +a^2 ) +C](https://www.tinkutara.com/question/Q179674.png)
$${b}\int\mathrm{arctan}\:\frac{{a}}{{x}}\:{dx}= \\ $$$$\:\:\:\:\:\begin{bmatrix}{{u}'=\mathrm{1}\:\rightarrow\:{u}={x}}\\{{v}=\mathrm{arctan}\:\frac{{x}}{{a}}\:\rightarrow\:{v}'=β\frac{{a}}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }}\end{bmatrix} \\ $$$$={bx}\mathrm{arctan}\:\frac{{x}}{{a}}\:+{ab}\int\frac{{x}}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{dx}= \\ $$$$={bx}\mathrm{arctan}\:\frac{{x}}{{a}}\:+\frac{{ab}}{\mathrm{2}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)\:+{C} \\ $$