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Question Number 44498 by peter frank last updated on 30/Sep/18

Commented by maxmathsup by imad last updated on 30/Sep/18

$$changement{4}^x=tgive{e}^{xln\left(4\right)}=t\Rightarrow xln\left(4\right)=ln\left(t\right)\Rightarrow dx=\frac{1}{tln\left(4\right)}dt$$$$I=\int 4^{4^t}.4^{t}t\frac{dt}{tln\left(4\right)}=\frac{1}{ln\left(4\right)}\int 4^t.4^{4^t}dtalsochangement{4}^t=ugive$$$$dt=\frac{du}{uln\left(4\right)}\Rightarrow I=\frac{1}{ln\left(4\right)}\int u.4^u\frac{du}{uln\left(4\right)}=\frac{1}{ln\left(4\right)}\int 4^{u}du$$$$=\frac{1}{ln{\left(4\right)}^2}\int e^{uln\left(4\right)}du=\frac{1}{{\left(ln\left(4\right)\right)}^3}{4}^u=\frac{1}{{\left(ln\left(4\right)\right)}^3}{4}^{4^t}=\frac{1}{{\left(ln\left(4\right)\right)}^3}{4}^{4^{4^x}}+c$$$$perhapsthereissomethingwronginthequestion..!$$

Commented by peter frank last updated on 30/Sep/18

$$\mathrm{true}\mathrm{sir}\mathrm{its}\mathrm{typing}\mathrm{error}\frac{}{{\left({\log }_{\mathrm{e}}4\right)}^3}$$

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