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I am helping my teenager son. So far I have been able to answer all his questions, and to solve all the problems he has had problems with. Except this one.

$\int{{3^x}{e^x}dx}$

Please, help me keep my success track and my son's confidence.

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Hint: try rewriting $3^x$ as $e^{x \ln 3}$.

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  • $\begingroup$ I see! (why didn't see it before, it's been two days of having this in my mind) thank you very much! $\endgroup$ – PA. Dec 16 '10 at 13:41
  • $\begingroup$ @PA01: Even without knowing this identity, there are ways. Did you try integration by parts? $\endgroup$ – Aryabhata Dec 16 '10 at 14:40
  • $\begingroup$ yes I did, I tried both u=e^x and v=e^x . To my despair, both lead me to integrals of similar type e^x k^x. I had obviously overlooked some basic exponential techniques. $\endgroup$ – PA. Dec 16 '10 at 14:58
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    $\begingroup$ @PA01: Leading to the same can sometimes be a bonus! If $\displaystyle I = \int 3^x e^x \text{dx}$ then, $\displaystyle I = \int 3^x \dfrac{d(e^x)}{dx} \text{dx} = 3^x e^x - \int \dfrac{d (3^x)}{dx} e^x \text{dx} = 3^x e^x - \log 3 \int 3^x e^x \text{dx}$ Thus $\displaystyle I = 3^x e^x - \log 3 \times I$ Hence $\displaystyle I = \dfrac{3^x e^x}{1 + \log 3} + \text{constant}$ $\endgroup$ – Aryabhata Dec 16 '10 at 15:04
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    $\begingroup$ I remember a similar trick being used for integrals of exp(ax)sin(bx) because differentiating the sine twice brings you back. $\endgroup$ – Ross Millikan Dec 16 '10 at 15:15
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Why not simply $\int 3^x e^x \mathrm dx = \int (3e)^x \mathrm dx = \frac{(3e)^x}{\ln(3e)} + Constant$

To see that $\int (3e)^x \mathrm dx = \frac{(3e)^x}{\ln(3e)}+C$, You can differentiate $(3e)^x$ i.e. Let

$$y = (3e)^x$$

$$\ln y = x \ln(3e)$$

$$ \mathrm dy/\mathrm dx = y \ln(3e) = (3e)^x \ln(3e)$$

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    $\begingroup$ Can you explain why it is not? I do not get it. I must be overlooking something easy $\endgroup$ – picakhu Dec 16 '10 at 16:13
  • $\begingroup$ Sorry, I was sleepy as I wrote that one. You are right. (I deleted the erroneous comment.) $\endgroup$ – J. M. is a poor mathematician Dec 17 '10 at 0:48
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    $\begingroup$ @J.M., deleting a comment on which other comments depend leads, more often than not, to more confusion instead of less. Because now picakhu's comment (and any further comment) is erroneous. $\endgroup$ – I. J. Kennedy Feb 24 '11 at 15:59

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