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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

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

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

$$\mathrm{let}\mathrm{u}=4^{\mathrm{x}}+1\Rightarrow \mathrm{du}=4^{\mathrm{x}}\ln 4\mathrm{dx}$$$$\int \frac{4^{\mathrm{x}}}{4^{\mathrm{x}}+1}\mathrm{dx}=\frac{1}{\ln 4}\int \frac{\mathrm{du}}{\mathrm{u}}=\frac{\ln \mid \mathrm{u}\mid }{\ln 4}=\frac{\ln (4^{\mathrm{x}}+1)}{\ln 4}={\log }_4(4^{\mathrm{x}}+1)+\mathrm{C}$$

Answered by Mathspace last updated on 30/Jun/22

$$I=\int \frac{4^x}{1+4^x}dxwedothechangement$$$$4^x=t\Rightarrow e^{xln\left(4\right)}=t\Rightarrow xln4=lnt\Rightarrow$$$$x=\frac{lnt}{2ln\left(2\right)}\Rightarrow$$$$I=\int \frac{t}{t+1}\frac{dt}{2tln2}$$$$=\frac{1}{2ln2}\int \frac{dt}{t+1}=\frac{1}{2ln2}ln\mid t+1\mid +c$$$$=\frac{1}{2ln2}ln(1+4^x)+c$$

Commented by CElcedricjunior last updated on 30/Jun/22

$$\mathit{o}\mathit{u}$$$$=\mathit{x}-\frac{1}{\mathbf{ln}2}\mathbf{ln}\mid \frac{2^{\mathit{x}}}{\sqrt{4^{\mathit{x}}+1}}\mid +\mathit{c}$$$$$$$$$$$$......\mathit{l}\mathit{e}\mathit{c}\acute{\mathit{e}\mathrm{l}}\stackrel{`}{\mathbf{e}\mathit{b}\mathit{r}\mathit{e}}\mathbf{cedric}\mathbf{junior}..........$$

Answered by CElcedricjunior last updated on 30/Jun/22

$$\int \frac{4^{\mathit{x}}}{4^{\mathit{x}}+1}\mathit{d}\mathit{x}=\frac{1}{\mathit{l}\mathit{n}4}\int \frac{\mathit{l}\mathit{n}4{\mathit{e}}^{\mathit{x}\mathit{l}\mathit{n}4}\mathit{d}\mathit{x}}{{\mathit{e}}^{\mathit{x}\mathit{l}\mathit{n}4}+1}$$$$=\frac{1}{\mathbf{ln}4}\mathbf{ln}\mid 4^{\mathit{x}}+1\mid +C\mathit{a}\mathit{v}\mathit{e}\mathit{c}\mathit{C}\in \mathbb{R}$$$$$$$$$$$$\mathcal{L}ec\acute{e}\mathit{l}\stackrel{`}{\mathit{e}\mathit{b}\mathit{r}\mathit{e}}\mathit{c}\mathit{e}\mathit{d}\mathit{r}\mathit{i}\mathit{c}\mathit{j}\mathit{u}\mathit{n}\mathit{i}\mathit{o}\mathit{r}$$$$$$

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