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.


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


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

  • $\begingroup$ 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

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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