The problem with your method is that you're using primitives and not definite integrals:
$\sum\limits_{n = 0}^\infty {\frac{1}{{n + 1}}{{\left( {\frac{x}{4}} \right)}^{n + 1}}} = - \log \left( {1 - \frac{x}{4}} \right) = - \log \left( {4 - x} \right) + \log 4 = \int\limits_0^x {\frac{1}{{4 - t}}dt} $
Since you know
$-\log \left( {1 - x} \right) = \sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{n}} $
You simply plug in $1/4$. You get
$\eqalign{ & - \frac{1}{2}\log \left( {1 - \frac{1}{4}} \right) = \frac{1}{2} \sum\limits_{n = 1}^\infty {\frac{1}{{n{4^n}}}} \cr & \frac{1}{2}\log \left( {\frac{4}{3}} \right) = \frac{1}{2}\sum\limits_{n = 1}^\infty {\frac{1}{{n{4^n}}}} \cr} $
You should be thinking about differentiating, not integrating. You have
$\frac{1}{2}\sum\limits_{n = 1}^\infty {n{{\left( {\frac{1}{4}} \right)}^n}} $
So you might want to find
$\frac{1}{2}\sum\limits_{n = 1}^\infty {n{x^n}} = f\left( x \right)$
Adn the plug in $1/4$. Use
\eqalign{ & F\left( x \right) = \frac{1}{{1 - x}} = \sum\limits_{n = 0}^\infty {{x^n}} \cr & x\frac{d}{{dx}}F\left( x \right) = x\frac{d}{{dx}}\frac{1}{{1 - x}} = x\frac{d}{{dx}}\sum\limits_{n = 0}^\infty {{x^n}} \cr & xF'\left( x \right) = \frac{x}{{{{\left( {1 - x} \right)}^2}}} = \sum\limits_{n = 1}^\infty {n{x^n}} \cr & \frac{1}{2}xF'\left( x \right) = \frac{1}{2}\frac{x}{{{{\left( {1 - x} \right)}^2}}} = \frac{1}{2}\sum\limits_{n = 1}^\infty {n{x^n}} \cr}
Now plug in $1/4$ to get
$\frac{1}{2}\frac{{\frac{1}{4}}}{{{{\left( {1 - \frac{1}{4}} \right)}^2}}} = \frac{1}{2}\sum\limits_{n = 1}^\infty {n{{\left( {\frac{1}{4}} \right)}^n}} $
$\frac{2}{9} = \frac{1}{2}\sum\limits_{n = 0}^\infty {n{{\left( {\frac{1}{4}} \right)}^n}} $