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Question Number 172556 by SANOGO last updated on 28/Jun/22
∫_ ln(1+x^2 )dx
$$\int_{} {ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right){dx} \\ $$
Answered by floor(10²Eta[1]) last updated on 28/Jun/22
u=ln(1+x^2 )⇒du=((2x)/(1+x^2 )) dv=dx⇒v=x ⇒∫ln(1+x^2 )dx=xln(1+x^2 )−∫((2x^2 )/(1+x^2 ))dx =xln(1+x^2 )−∫(2−(2/(1+x^2 )))dx =xln(1+x^2 )−2x+2arctg(x)+C
$$\mathrm{u}=\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\Rightarrow\mathrm{du}=\frac{\mathrm{2x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{dv}=\mathrm{dx}\Rightarrow\mathrm{v}=\mathrm{x} \\ $$$$\Rightarrow\int\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx}=\mathrm{xln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)−\int\frac{\mathrm{2x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\mathrm{xln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)−\int\left(\mathrm{2}−\frac{\mathrm{2}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right)\mathrm{dx} \\ $$$$=\mathrm{xln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)−\mathrm{2x}+\mathrm{2arctg}\left(\mathrm{x}\right)+\mathrm{C} \\ $$$$ \\ $$
Answered by floor(10²Eta[1]) last updated on 28/Jun/22
I=∫ln(ax^n +b)dx u=ln(ax^n +b)⇒du=((anx^(n−1) )/(ax^n +k))dx dv=dx⇒v=x ⇒I=xln(ax^n +b)−∫((anx^n )/(ax^n +b))dx anx^n =(ax^n +b)n−bn I=xln(ax^n +b)−∫(n−((bn)/(ax^n +b)))dx =xln(ax^n +b)−nx+bn∫(dx/(ax^n +b))+C the integral ∫(dx/(ax^n +b)) can be solved using methods like: partial fractions ostrogradski method
$$\mathrm{I}=\int\mathrm{ln}\left(\mathrm{ax}^{\mathrm{n}} +\mathrm{b}\right)\mathrm{dx} \\ $$$$\mathrm{u}=\mathrm{ln}\left(\mathrm{ax}^{\mathrm{n}} +\mathrm{b}\right)\Rightarrow\mathrm{du}=\frac{\mathrm{anx}^{\mathrm{n}−\mathrm{1}} }{\mathrm{ax}^{\mathrm{n}} +\mathrm{k}}\mathrm{dx} \\ $$$$\mathrm{dv}=\mathrm{dx}\Rightarrow\mathrm{v}=\mathrm{x} \\ $$$$\Rightarrow\mathrm{I}=\mathrm{xln}\left(\mathrm{ax}^{\mathrm{n}} +\mathrm{b}\right)−\int\frac{\mathrm{anx}^{\mathrm{n}} }{\mathrm{ax}^{\mathrm{n}} +\mathrm{b}}\mathrm{dx} \\ $$$$\mathrm{anx}^{\mathrm{n}} =\left(\mathrm{ax}^{\mathrm{n}} +\mathrm{b}\right)\mathrm{n}−\mathrm{bn} \\ $$$$\mathrm{I}=\mathrm{xln}\left(\mathrm{ax}^{\mathrm{n}} +\mathrm{b}\right)−\int\left(\mathrm{n}−\frac{\mathrm{bn}}{\mathrm{ax}^{\mathrm{n}} +\mathrm{b}}\right)\mathrm{dx} \\ $$$$=\mathrm{xln}\left(\mathrm{ax}^{\mathrm{n}} +\mathrm{b}\right)−\mathrm{nx}+\mathrm{bn}\int\frac{\mathrm{dx}}{\mathrm{ax}^{\mathrm{n}} +\mathrm{b}}+\mathrm{C} \\ $$$$ \\ $$$$\mathrm{the}\:\mathrm{integral}\:\int\frac{\mathrm{dx}}{\mathrm{ax}^{\mathrm{n}} +\mathrm{b}}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved}\: \\ $$$$\mathrm{using}\:\mathrm{methods}\:\mathrm{like}: \\ $$$$\mathrm{partial}\:\mathrm{fractions} \\ $$$$\mathrm{ostrogradski}\:\mathrm{method} \\ $$
Commented by Mikenice last updated on 29/Jun/22
please which textbook can i found ostrogradski method
$${please}\:{which}\:{textbook} \\ $$$${can}\:{i}\:{found}\:{ostrogradski}\:{method} \\ $$
Commented by floor(10²Eta[1]) last updated on 29/Jun/22
google
$$\mathrm{google} \\ $$
Commented by Mikenice last updated on 29/Jun/22
how sir
$${how}\:{sir} \\ $$
Commented by floor(10²Eta[1]) last updated on 29/Jun/22
you dont know how to use google? just search it
$$\mathrm{you}\:\mathrm{dont}\:\mathrm{know}\:\mathrm{how}\:\mathrm{to}\:\mathrm{use}\:\mathrm{google}?\:\mathrm{just}\:\mathrm{search}\:\mathrm{it} \\ $$
Answered by CElcedricjunior last updated on 30/Jun/22
∫ln(1+x^2 )dx=k posons { ((u=ln(1+x^2 ))),((v′=1)) :}=> { ((u′=((2x)/(1+x^2 )))),((v=x)) :} =xln(1+x^2 )−∫((2x^2 )/(1+x^2 ))dx k=xln(1+x^2 )−2x+2∫(dx/(1+x^2 )) k=xln(1+x^2 )−2x+2arctanx+c c∈R .....Le ce^� le^� bre cedric junior.......
$$\int\boldsymbol{\mathrm{ln}}\left(\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)\boldsymbol{\mathrm{dx}}=\boldsymbol{\mathrm{k}} \\ $$$$\boldsymbol{\mathrm{posons}}\:\begin{cases}{\boldsymbol{\mathrm{u}}=\boldsymbol{\mathrm{ln}}\left(\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)}\\{\boldsymbol{\mathrm{v}}'=\mathrm{1}}\end{cases}=>\begin{cases}{\boldsymbol{\mathrm{u}}'=\frac{\mathrm{2}\boldsymbol{\mathrm{x}}}{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }}\\{\boldsymbol{\mathrm{v}}=\boldsymbol{\mathrm{x}}}\end{cases} \\ $$$$=\boldsymbol{\mathrm{xln}}\left(\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)−\int\frac{\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} }{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{\mathrm{k}}=\boldsymbol{\mathrm{xln}}\left(\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)−\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{2}\int\frac{\boldsymbol{\mathrm{dx}}}{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} } \\ $$$$\boldsymbol{\mathrm{k}}=\boldsymbol{\mathrm{xln}}\left(\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)−\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{2}\boldsymbol{\mathrm{arctanx}}+\boldsymbol{\mathrm{c}}\:\:\:\:\boldsymbol{\mathrm{c}}\in\mathbb{R} \\ $$$$…..\mathscr{L}\boldsymbol{\mathrm{e}}\:\boldsymbol{\mathrm{c}}\acute {\boldsymbol{\mathrm{e}l}}\grave {\boldsymbol{\mathrm{e}bre}}\:\boldsymbol{\mathrm{cedric}}\:\boldsymbol{\mathrm{junior}}……. \\ $$

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