# Integration of $\int \sqrt{x\sqrt{2x}} dx$ and $\int 3^x e^x dx$

I try to evaluate this two integrals, but I don't know how to proceed:

i) $\int \sqrt{x\sqrt{2x}} dx = \int {2^{\frac{1}{4}}\cdot x^{\frac{3}{4}}}$

ii) $\int 3^x e^x dx$

What's the best way to evaluate them? Substitution or Intergration by parts?

Any hints are appreciated.

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Should ii) be as in the title or as in the question? –  mrf Sep 12 '12 at 7:53
as in the title, I edited it, thanks. –  ulead86 Sep 12 '12 at 7:58
You don't solve integrals, you evaluate them. –  Stefan Smith Sep 12 '12 at 14:03

$$\sqrt{x\sqrt{2x}}=\left(x(2x)^{1/2}\right)^{1/2}=\left(x\cdot2^{1/2}x^{1/2}\right)^{1/2}=\left(2^{1/2}x^{3/2}\right)^{1/2}=2^{1/4}x^{3/4}\;.$$
Now you have $\displaystyle\int2^{1/4}x^{3/4}~dx=2^{1/4}\int x^{3/4}~dx$, which is just a power rule integration.
In the second problem, use the fact that $3^xe^x=(3e)^x$; I’m sure that you’ve been shown how to integrate $a^x$ for a constant $a$.
Thanks a lot, I edited the mistake. And yes, I know how to integrate $a^x$. Thanks for the answer. –  ulead86 Sep 12 '12 at 7:59