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the problem: $\int 3^{x^2}\, \mathrm{d}x$

I have tried doing some subitution but I can't seem to get a du that works.

$\sqrt{u} = 3^x$ which didn't work,

$u = 3^{x^2}$ -I couldn't find a du for this that I trust.

$\ln u= x^2 \ln 3$ and and several others.

I haven't found any way to rearrage the variables to get a nice du so I can solve it. Meaning I haven't found a way to account for all the $x$ inorder to do subutition.

Am I on the correct path, is subution the method to solve this?

The answer sheets says $$\frac{3^{x^2}}{\ln 9} + C$$

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Is it $3^{x^2}$ or $3^x$? – Mhenni Benghorbal Dec 9 '12 at 9:51
See wolframalpha – Karolis Juodelė Dec 9 '12 at 9:55
The derivative of $\dfrac{3^{x^2}}{\ln9}$ is $x3^{x^2}$. Maybe there is a typo in the problem. – P.. Dec 9 '12 at 10:00
Everybody cool down. What he's asking is to find the integral of $(3^x)^2 = 3^{2x} = 9^x$, which happily does have an antiderivative in elementary functions. – andybenji Dec 9 '12 at 10:07
@andybenji: This is what I already put in the comment. – Mhenni Benghorbal Dec 9 '12 at 10:08
up vote 3 down vote accepted

What the question is asking is for the antiderivative of $(3^x)^2 = 3^{2x} = 9^x$. Since $9 = e^{\ln(9)}$, we may substitute this in, so we have that $$ \int 9^x dx = \int e^{\ln(9)x} dx = \frac{e^{\ln(9)x}}{\ln(9)} + C = \frac{9^x}{\ln(9)} + C$$

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If$\; C=(3^x)^2=3^{2x}$. We could take logarithm both sides.

Hence, $$ \ln C=\ln 3^{2x}=2x\ln 3\\ \begin{align} C=e^{2x\ln 3}\\ \therefore\; 3^{2x}=e^{2x\ln 3} \end{align} $$ Therefore,using substitution $ u=2x\ln3 $ $$ \int 3^{2x}dx\; \begin{align} =\int e^{2x\ln3}dx =\frac{1}{2\ln3}\int e^udu =\frac{e^{2x\ln3}}{2\ln3}+C =\frac{(3^x)^2}{2\ln3}+C \end{align} $$

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See Error function. Your problem reduces to this by trivial rescaling: $3^{x^2} = e^{-(i(\ln 3)^{1/2} x)^2}$.

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The identity is wrong and I fail to see how it would help. – Did Dec 19 '12 at 17:46
@did: Corrected, thanks. If the OP really meant not $3^{2x}$ but $3^{x^2}$, as she stated very explicitly by writing the braces, the answer would be $\frac{\sqrt \pi}{2 i \sqrt{\ln 3}} \mathrm{erf[i \sqrt{\ln 3} \cdot x]}$. – Alexander Shamov Dec 20 '12 at 20:41

I think you are asking the following

$$ \int 3^{2x}dx = \int 9^x dx = \int e^{\ln(9) x} dx = \frac{e^{\ln(9) x}}{\ln(9)}=\frac{{3^{2x}}}{\ln(9)}. $$

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