Non-inverting-Amp bias current: understanding output error
Interesting, because when I use a Superposition Theorem for this circuit 



The equation for \$V_o\$ 
if I left \$I_O\$ alone (\$V_{IN}\$ is short and \$I_B\$ are open) is:  

$$V_O = I_O*R_N$$ 

for \$I_B\$ we have: 

$$ V_O = I_BR_N - I_BR_g\left(\frac{R_N}{R_1}+1\right) = I_B\left(R_N - \frac{R_g(R_1+R_N)}{R_1}\right)$$ 

So finally we have: 

$$ V_O = \left(\frac{R_N}{R_1}+1\right) V_{IN}+I_B\left(R_N - \frac{R_g(R_1 + R_N)}{R_1}\right)+I_OR_N$$ 

What do you think? 

EDIT

After some thought about this I get to this conclusion: 

\$I_B = \frac{I_P+I_N}{2}\$ 

\$I_O = |I_P+I_N| \$ 

Where \$I_P\$ is a Non-Inverting input bias current and \$I_N\$ - Inverting Input bias current. 

From this I get: 

\$ I_P = I_B-0.5I_O \$

and 

\$ I_N = I_B+0.5I_O \$ 

And the circuit diagram will look like this:



And for this circuit this equation is correct ( from Tietze, Schenk: Electronic circuits) 

$$V_o = \left(1+\frac{R_N}{R_1}\right)V_i + I_B\left(R_N - \frac{R_g(R_1 + R_N)}{R_1}\right) + \frac{I_o}{2} \left(R_N + \frac{R_g(R_1 + R_N)}{R_1}\right)$$

Try superposition and noticed that the \$V_P\$ voltage is equal to:

\$V_p=-(I_B-\frac{I_o}{2})R_g + V_i\$  

EDIT 2

Let us try to find \$V_o\$ for these two cases 


The voltage at \$V_p\$ node is 

$$ V_p = -I_B*R_g$$

and the output voltage is \$V_O' = V_p*A_v\$ and the Non-Inverting gain is \$A_v = (1+\frac{R_N}{R_1}) = \frac{R_1+R_N}{R_1}\$ therefore

$$V_O' =  -I_B*R_g*\frac{R_1+R_N}{R_1}=-I_B \frac{R_g(R_1+R_N)}{R_1} $$ 

And for the second case, we have 

$$V_O'' = I_B*R_N$$ 

And finally 

$$V_O = V_O'+V_O'' = -I_B \frac{R_g(R_1+R_N)}{R_1} + I_BR_N = I_BR_N -I_B \frac{R_g(R_1+R_N)}{R_1} = $$
$$=I_B\left(R_N - \frac{R_g(R_1 + R_N)}{R_1}\right) $$

If we repeat this for \$I_o\$ current we will get the plus term because now \$V_p\$ voltage is positive.