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Question Number 21772 by j.masanja06@gmail.com last updated on 03/Oct/17

integrate ∫2^(4x) dx

$${integrate} \\ $$$$\int\mathrm{2}^{\mathrm{4}{x}} {dx} \\ $$

Answered by mrW1 last updated on 03/Oct/17

∫2^(4x) dx =(1/4)∫2^(4x) d(4x) =(1/4)∫2^t dt with t=4x =(1/4)∫e^(tln 2) dt =(1/(4ln 2))∫e^(tln 2) d(tln 2) =(1/(4ln 2))∫e^u du with u=tln 2=ln 2^t =ln 2^(4x) =(1/(4ln 2))×e^u +C =(1/(4ln 2))×e^(ln 2^(4x) ) +C =(2^(4x) /(4ln 2))+C

$$\int\mathrm{2}^{\mathrm{4x}} \mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int\mathrm{2}^{\mathrm{4x}} \mathrm{d}\left(\mathrm{4x}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int\mathrm{2}^{\mathrm{t}} \mathrm{dt}\:\:\mathrm{with}\:\mathrm{t}=\mathrm{4x} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int\mathrm{e}^{\mathrm{tln}\:\mathrm{2}} \:\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4ln}\:\mathrm{2}}\int\mathrm{e}^{\mathrm{tln}\:\mathrm{2}} \:\mathrm{d}\left(\mathrm{tln}\:\mathrm{2}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4ln}\:\mathrm{2}}\int\mathrm{e}^{\mathrm{u}} \:\mathrm{du}\:\:\mathrm{with}\:\mathrm{u}=\mathrm{tln}\:\mathrm{2}=\mathrm{ln}\:\mathrm{2}^{\mathrm{t}} =\mathrm{ln}\:\mathrm{2}^{\mathrm{4x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4ln}\:\mathrm{2}}×\mathrm{e}^{\mathrm{u}} +\mathrm{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4ln}\:\mathrm{2}}×\mathrm{e}^{\mathrm{ln}\:\mathrm{2}^{\mathrm{4x}} } +\mathrm{C} \\ $$$$=\frac{\mathrm{2}^{\mathrm{4x}} }{\mathrm{4ln}\:\mathrm{2}}+\mathrm{C} \\ $$

Answered by sma3l2996 last updated on 03/Oct/17

=∫e^(4xln2) dx=(2^(4x) /(4ln2))+C

$$=\int{e}^{\mathrm{4}{xln}\mathrm{2}} {dx}=\frac{\mathrm{2}^{\mathrm{4}{x}} }{\mathrm{4}{ln}\mathrm{2}}+{C} \\ $$

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