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Question Number 67937 by A8;15: last updated on 02/Sep/19

Commented by mathmax by abdo last updated on 02/Sep/19

at form of serie  I =∫(√e^x )dx = ∫  e^(x/2) dx =∫(Σ_(n=0) ^∞   ((((x/2))^n )/(n!)))dx  =Σ_(n=0) ^∞  (1/(n!2^n )) ∫  x^n  dx = Σ_(n=0) ^∞    (x^(n+1) /((n+1)n! 2^n ))   +C   and  this  serie converges unif. with radius infinite.

$${at}\:{form}\:{of}\:{serie} \\ $$$${I}\:=\int\sqrt{{e}^{{x}} }{dx}\:=\:\int\:\:{e}^{\frac{{x}}{\mathrm{2}}} {dx}\:=\int\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(\frac{{x}}{\mathrm{2}}\right)^{{n}} }{{n}!}\right){dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{{n}!\mathrm{2}^{{n}} }\:\int\:\:{x}^{{n}} \:{dx}\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{{x}^{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right){n}!\:\mathrm{2}^{{n}} }\:\:\:+{C}\:\:\:{and}\:\:{this} \\ $$$${serie}\:{converges}\:{unif}.\:{with}\:{radius}\:{infinite}. \\ $$

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