prove-that-2-ln-x-dx-x-2-ln-x-ln-xe-C-

Question Number 44512 by arvinddayama01@gmail.com last updated on 30/Sep/18

p r o v e t h a t : - \int 2^{\ln x} dx = \frac{x . 2^{\ln x}}{\ln (xe)} + C

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

l e t I = \int 2^{l n (x)} d x c h a 7 g e m e n t l n (x) = t g i v e

I = \int 2^{t} e^{t} d t = \int e^{t l n (2) + t} d t = \int e^{(1 + l n (2)) t} d t

= \frac{1}{1 + l n (2)} e^{(1 + l n (2)) t} + c = \frac{1}{1 + l n (2)} e^{(1 + l n (2)) l n (x)} + c

= \frac{1}{1 + l n (2)} x . 2^{l n (x)} + c s o t h e r e i s a e r r o r a t t h e q u e s t i o n \dots!

Answered by MJS last updated on 01/Oct/18

2^{\ln x} = {(e^{\ln 2})}^{\ln x} = e^{\ln 2 \times \ln x} = {(e^{\ln x})}^{\ln 2} = x^{\ln 2}

- \int x^{\ln 2} d x = - \frac{1}{1 + \ln 2} x^{1 + \ln 2} = - \frac{x \times x^{\ln 2}}{1 + \ln 2} =

= - \frac{x \times 2^{\ln x}}{\ln e + \ln 2} = - \frac{x \times 2^{lm x}}{\ln 2 e}

so {it}^{'} s wrong

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