Question-220715

Question Number 220715 by Noorzai last updated on 18/May/25

Commented by Noorzai last updated on 18/May/25

$${do}\:{you}\:{help}\:{me} \\ $$$$ \\ $$

Answered by SdC355 last updated on 18/May/25

∫ xa^x dx= ((d )/dx)(xa^x )=a^x +x∙ln(a)a^x ∫ ((d )/dx)(x∙a^x ) dx=∫ d(x∙a^x )=∫ a^x dx+∫ xln(a)a^x dx x∙a^x +Const=∫ a^x dx+ln(a)∫ xa^x dx xa^x −∫ a^x dx+Const xa^x −(a^x /(ln(a)))+Const=ln(a)∫ xa^x dx (((x∙ln(a)−1)a^x )/(ln^2 (a)))+Const=∫ xa^x dx ∴ ∫ 3x∙5^x dx=((3(x∙ln(5)−1)5^x )/(ln^2 (5)))+Const

$$\int\:\:{xa}^{{x}} \:\mathrm{d}{x}= \\ $$$$\frac{\mathrm{d}\:\:}{\mathrm{d}{x}}\left({xa}^{{x}} \right)={a}^{{x}} +{x}\centerdot\mathrm{ln}\left({a}\right){a}^{{x}} \\ $$$$\int\:\:\frac{\mathrm{d}\:\:}{\mathrm{d}{x}}\left({x}\centerdot{a}^{{x}} \right)\:\mathrm{d}{x}=\int\:\:\mathrm{d}\left({x}\centerdot{a}^{{x}} \right)=\int\:\:{a}^{{x}} \:\mathrm{d}{x}+\int\:\:{x}\mathrm{ln}\left({a}\right){a}^{{x}} \:\mathrm{d}{x} \\ $$$${x}\centerdot{a}^{{x}} +\mathrm{Const}=\int\:\:{a}^{{x}} \:\mathrm{d}{x}+\mathrm{ln}\left({a}\right)\int\:\:{xa}^{{x}} \:\mathrm{d}{x} \\ $$$${xa}^{{x}} −\int\:{a}^{{x}} \:\mathrm{d}{x}+\mathrm{Const} \\ $$$${xa}^{{x}} −\frac{{a}^{{x}} }{\mathrm{ln}\left({a}\right)}+\mathrm{Const}=\mathrm{ln}\left({a}\right)\int\:{xa}^{{x}} \:\mathrm{d}{x} \\ $$$$\frac{\left({x}\centerdot\mathrm{ln}\left({a}\right)−\mathrm{1}\right){a}^{{x}} }{\mathrm{ln}^{\mathrm{2}} \left({a}\right)}+\mathrm{Const}=\int\:\:{xa}^{{x}} \:\mathrm{d}{x} \\ $$$$\therefore\:\int\:\:\mathrm{3}{x}\centerdot\mathrm{5}^{{x}} \:\mathrm{d}{x}=\frac{\mathrm{3}\left({x}\centerdot\mathrm{ln}\left(\mathrm{5}\right)−\mathrm{1}\right)\mathrm{5}^{{x}} }{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{5}\right)}+\mathrm{Const} \\ $$

Commented by Noorzai last updated on 18/May/25

$${thanks}\:{can}\:{you}\:{cleaer} \\ $$

Answered by mr W last updated on 18/May/25

=3∫xe^(xln 5) dx =(3/(ln 5))∫xd(e^(xln 5) ) =(3/(ln 5))(xe^(xln x) −∫e^(xln 5) dx) =(3/(ln 5))(x5^x −(1/(ln 5))∫e^(xln 5) d(xln 5)) =(3/(ln 5))(x5^x −(e^(xln 5) /(ln 5)))+C =(3/(ln 5))(x5^x −(5^x /(ln 5)))+C =(3/(ln 5))(x−(1/(ln 5)))5^x +C

$$=\mathrm{3}\int{xe}^{{x}\mathrm{ln}\:\mathrm{5}} {dx} \\ $$$$=\frac{\mathrm{3}}{\mathrm{ln}\:\mathrm{5}}\int{xd}\left({e}^{{x}\mathrm{ln}\:\mathrm{5}} \right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{ln}\:\mathrm{5}}\left({xe}^{{x}\mathrm{ln}\:{x}} −\int{e}^{{x}\mathrm{ln}\:\mathrm{5}} {dx}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{ln}\:\mathrm{5}}\left({x}\mathrm{5}^{{x}} −\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{5}}\int{e}^{{x}\mathrm{ln}\:\mathrm{5}} {d}\left({x}\mathrm{ln}\:\mathrm{5}\right)\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{ln}\:\mathrm{5}}\left({x}\mathrm{5}^{{x}} −\frac{{e}^{{x}\mathrm{ln}\:\mathrm{5}} }{\mathrm{ln}\:\mathrm{5}}\right)+{C} \\ $$$$=\frac{\mathrm{3}}{\mathrm{ln}\:\mathrm{5}}\left({x}\mathrm{5}^{{x}} −\frac{\mathrm{5}^{{x}} }{\mathrm{ln}\:\mathrm{5}}\right)+{C} \\ $$$$=\frac{\mathrm{3}}{\mathrm{ln}\:\mathrm{5}}\left({x}−\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{5}}\right)\mathrm{5}^{{x}} +{C} \\ $$

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