Why feedback resistor be at least 10 times of the input resistance?
It should be much larger than the input resistor.  The number 10 is just a nice round number somebody picked.
The feedback resistor is a necessary evil.  Ideally you wouldn't want to have it, but without it the output is likely to saturate.  If the op-amp saturates then the op-amp will no longer behave like an op-amp and you won't get anything close to the integral at the output.
The feedback resistor takes some current that would otherwise be going into the capacitor to create the integral function.  Having it there creates a little bit of error in the output.  The larger the feedback resistor is, the less error there will be.
Normally if you are trying to get a pure integral function, then you make a trade-off between having the feedback resistor as large as possible, but not so large that the output saturates.
So how does it prevent the output from saturating?
Imagine you put a sine-wave into the input that is centered around 0V.  Theoretically the average value of this waveform is 0V, so the integral should be bounded.  But in reality the average won't be exactly 0V, and even if it was, the op-amp input has finite input resistance and will take some bias current.
Without the feedback resistor the output would eventually wander all the way to one of the power supply voltages and the integrator would stop working.  What the feedback resistor does is steal a little current from the output in such a way that it tends to very slowly pull the integral back towards 0V.  So as long as the average of the input is very close to 0V then the output wont wander too far.
Note that if your capacitor is leaky enough, you may not need to add a separate resistor, since you essentially have one inside the capacitor.