The sum of an infinite series with integral $1+\dfrac{1}{9}+\dfrac{1}{45}+\dfrac{1}{189}+\dfrac{1}{729}+\dots=\sum\limits_{n=1}^\infty \dfrac{1}{(2n-1)\cdot 3^{n-1}}$ 
I got:
$\sum\limits_{n=1}^\infty \dfrac{1}{(2n-1)\cdot 3^{n-1}}=\sum\limits_{n=1}^\infty \dfrac{\int\limits_0^1 x^{2n-2}\,dx}{3^{n-1}}=\dots$
And no idea how to impove.
Thx!

After writing $\frac{1}{2n-1}$ as an integral, you have the series
$$\sum_{n=1}^\infty \int_0^1 \left(\frac{x^2}{3}\right)^{n-1}\,dx.$$
Since the geometric series
$$\sum_{k=0}^\infty \left(\frac{x^2}{3}\right)^k$$
converges uniformly on the interval $[0,1]$, we can interchange summation and integration, and obtain
$$\sum_{n=1}^\infty \frac{1}{(2n-1)3^{(n-1)}} = \int_0^1 \frac{3}{3-x^2}\,dx.$$
The integral can be evaluated in different ways. With partial fraction decomposition
$$\frac{3}{3-x^2} = \frac{\sqrt{3}}{2}\left(\frac{1}{\sqrt{3}-x} + \frac{1}{\sqrt{3}+x}\right)$$
we get
$$\int_0^1 \frac{3}{3-x^2}\,dx = \frac{\sqrt{3}}{2}\left[\log (\sqrt{3}+x) - \log (\sqrt{3}-x)\right]_0^1 = \frac{\sqrt{3}}{2}\log \frac{\sqrt{3}+1}{\sqrt{3}-1} = \frac{\sqrt{3}}{2}\log (2+\sqrt{3}).$$
