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Question Number 172635 by Mikenice last updated on 29/Jun/22

Answered by MJS_new last updated on 29/Jun/22

∫(4^x /(4^x +1))dx= [t=4^x +1 → dx=(dt/(4^x ln 4))] =(1/(ln 4))∫(dt/t)=((ln t)/(ln 4))= =((ln (4^x +1))/(ln 4))+C

$$\int\frac{\mathrm{4}^{{x}} }{\mathrm{4}^{{x}} +\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{4}^{{x}} +\mathrm{1}\:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{4}^{{x}} \:\mathrm{ln}\:\mathrm{4}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{4}}\int\frac{{dt}}{{t}}=\frac{\mathrm{ln}\:{t}}{\mathrm{ln}\:\mathrm{4}}= \\ $$$$=\frac{\mathrm{ln}\:\left(\mathrm{4}^{{x}} +\mathrm{1}\right)}{\mathrm{ln}\:\mathrm{4}}+\mathrm{C} \\ $$

Answered by floor(10²Eta[1]) last updated on 29/Jun/22

let u=4^x +1⇒du=4^x ln4dx ∫(4^x /(4^x +1))dx=(1/(ln4))∫(du/u)=((ln∣u∣)/(ln4))=((ln(4^x +1))/(ln4))=log_4 (4^x +1)+C

$$\mathrm{let}\:\mathrm{u}=\mathrm{4}^{\mathrm{x}} +\mathrm{1}\Rightarrow\mathrm{du}=\mathrm{4}^{\mathrm{x}} \mathrm{ln4dx} \\ $$$$\int\frac{\mathrm{4}^{\mathrm{x}} }{\mathrm{4}^{\mathrm{x}} +\mathrm{1}}\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{ln4}}\int\frac{\mathrm{du}}{\mathrm{u}}=\frac{\mathrm{ln}\mid\mathrm{u}\mid}{\mathrm{ln4}}=\frac{\mathrm{ln}\left(\mathrm{4}^{\mathrm{x}} +\mathrm{1}\right)}{\mathrm{ln4}}=\mathrm{log}_{\mathrm{4}} \left(\mathrm{4}^{\mathrm{x}} +\mathrm{1}\right)+\mathrm{C} \\ $$

Answered by Mathspace last updated on 30/Jun/22

I=∫ (4^x /(1+4^x ))dx we do the changement 4^x =t ⇒e^(xln(4)) =t ⇒xln4=lnt ⇒ x=((lnt)/(2ln(2))) ⇒ I=∫ (t/(t+1))(dt/(2tln2)) =(1/(2ln2))∫(dt/(t+1))=(1/(2ln2))ln∣t+1∣ +c =(1/(2ln2))ln(1+4^x )+c

$${I}=\int\:\frac{\mathrm{4}^{{x}} }{\mathrm{1}+\mathrm{4}^{{x}} }{dx}\:\:{we}\:{do}\:{the}\:{changement} \\ $$$$\mathrm{4}^{{x}} ={t}\:\Rightarrow{e}^{{xln}\left(\mathrm{4}\right)} ={t}\:\Rightarrow{xln}\mathrm{4}={lnt}\:\Rightarrow \\ $$$${x}=\frac{{lnt}}{\mathrm{2}{ln}\left(\mathrm{2}\right)}\:\Rightarrow \\ $$$${I}=\int\:\:\frac{{t}}{{t}+\mathrm{1}}\frac{{dt}}{\mathrm{2}{tln}\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{ln}\mathrm{2}}\int\frac{{dt}}{{t}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}{ln}\mathrm{2}}{ln}\mid{t}+\mathrm{1}\mid\:+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{ln}\mathrm{2}}{ln}\left(\mathrm{1}+\mathrm{4}^{{x}} \right)+{c} \\ $$

Commented by CElcedricjunior last updated on 30/Jun/22

ou =x−(1/(ln2))ln∣(2^x /( (√(4^x +1))))∣+c ......le ce^� le^� bre cedric junior..........

$$\boldsymbol{{ou}} \\ $$$$=\boldsymbol{{x}}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{ln}}\mathrm{2}}\boldsymbol{\mathrm{ln}}\mid\frac{\mathrm{2}^{\boldsymbol{{x}}} }{\:\sqrt{\mathrm{4}^{\boldsymbol{{x}}} +\mathrm{1}}}\mid+\boldsymbol{{c}} \\ $$$$ \\ $$$$ \\ $$$$......\boldsymbol{{le}}\:\boldsymbol{{c}}\acute {\boldsymbol{{e}}\mathrm{l}}\grave {\boldsymbol{\mathrm{e}bre}}\:\boldsymbol{\mathrm{cedric}}\:\boldsymbol{\mathrm{junior}}.......... \\ $$

Answered by CElcedricjunior last updated on 30/Jun/22

∫(4^x /(4^x +1))dx=(1/(ln4))∫((ln4e^(xln4) dx)/(e^(xln4) +1)) =(1/(ln4))ln∣4^x +1∣+C avec C∈R Le ce^� le^� bre cedric junior

$$\int\frac{\mathrm{4}^{\boldsymbol{{x}}} }{\mathrm{4}^{\boldsymbol{{x}}} +\mathrm{1}}\boldsymbol{{dx}}=\frac{\mathrm{1}}{\boldsymbol{{ln}}\mathrm{4}}\int\frac{\boldsymbol{{ln}}\mathrm{4}\boldsymbol{{e}}^{\boldsymbol{{xln}}\mathrm{4}} \boldsymbol{{dx}}}{\boldsymbol{{e}}^{\boldsymbol{{xln}}\mathrm{4}} +\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\boldsymbol{\mathrm{ln}}\mathrm{4}}\boldsymbol{\mathrm{ln}}\mid\mathrm{4}^{\boldsymbol{{x}}} +\mathrm{1}\mid+{C}\:\boldsymbol{{avec}}\:\boldsymbol{{C}}\in\mathbb{R} \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\mathscr{L}{e}\:{c}\acute {{e}}\boldsymbol{{l}}\grave {\boldsymbol{{e}bre}}\:\boldsymbol{{cedric}}\:\boldsymbol{{junior}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

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