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Question Number 87833 by M±th+et£s last updated on 06/Apr/20

$$1)find\int 2^{ln\left(x\right)}dx$$$$2)prove\sqrt[i]{i}=reslnumber$$

Commented by mr W last updated on 06/Apr/20

$$2)$$$$i=e^{\frac{i\pi }{2}}$$$$\sqrt[i]{i}=i^{\frac{1}{i}}={\left(e^{\frac{i\pi }{2}}\right)}^{\frac{1}{i}}=e^{\frac{\pi }{2}}=\sqrt{e^{\pi }}=real$$

Commented by john santu last updated on 06/Apr/20

$$1)\mathrm{by}\mathrm{parts}$$$$\mathrm{u}=2^{\ln \mathrm{x}}\Rightarrow \mathrm{du}=2^{\ln \mathrm{x}}.\frac{\ln 2}{\mathrm{x}}\mathrm{dx}$$$$\mathrm{I}=\mathrm{x}.2^{\ln \mathrm{x}}-\ln 2\int 2^{\ln \mathrm{x}}\mathrm{dx}$$$$(1+\ln 2)\mathrm{I}=\mathrm{x}.2^{\ln \mathrm{x}}$$$$\mathrm{I}=\frac{\mathrm{x}.2^{\ln \mathrm{x}}}{\ln \left(2\mathrm{e}\right)}+\mathrm{c}$$

Commented by M±th+et£s last updated on 06/Apr/20

$$thankyou$$

Commented by M±th+et£s last updated on 06/Apr/20

$$tankyou$$

Commented by mathmax by abdo last updated on 06/Apr/20

$$2){}^i\sqrt{i}=i^{\frac{1}{i}}=i^{-i}={\left(e^{\frac{i\pi }{2}}\right)}^{-i}=e^{\frac{\pi }{2}}and{e}^{\frac{\pi }{2}}\in R$$

Answered by petrochengula last updated on 06/Apr/20

$$\int 2^{lnx}dx$$$$e^{ln2}=2then{2}^{lnx}={\left(e^{ln2}\right)}^{lnx}={\left(2^{lnx}\right)}^{ln2}=x^{ln2}$$$$\int 2^{lnx}dx=\int x^{ln2}dx=\frac{x^{ln2+1}}{ln2+1}+C$$

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