Op Amp Integrator Circuit - Why is there a linear current and voltage for a short time before exponential values?
Opamp tries to keep both of its inputs at the same voltage. Since the positive input is fixed at ground, the inverting input must be at 0V too. The inputs of an opamp are very high impedance, so we can assume that no current can flow in or out of them. The only way the current can flow is into the output of the opamp.

Since V1 is connected to the inverting input trough a 1K resistor, the current trough that 1K resistor has to be 5mA to result in the inverting input voltage at 0V, as such, the opamp output becomes a 5mA current sink, charging the capacitor with the constant current. 

When you connect a capacitor to a constant current source, the voltage is linear with time. So this explains the linear portion of the graph.

After that, the opamp can no longer provide output negative enough (the supply rails are probably +-15V) to sustain the constant current, the current decreases and the voltage graph flattens out. Ideally, the capacitor would charge linearly forever.