Active loads increase the gain of an amplifier, since they present a high impedance in AC. But they can draw fairly large currents in DC. A simple resistor would need to have a large value for high AC gain, but this large value would make it hard or even impossible to obtain the required DC current, because the required voltages can become very high. Let's assume you need a current of 1mA in each transistor and you need a load resistance of 1MΩ. That means you need 1000V across that resistor!
Most opamps have a single-ended output, but a differential input. So, at some point a conversion is needed from differential to single-ended.
With a single resistor, like in the first figure, the gain of the differential stage is halved, since the left transistor does not contribute anything to the output voltage, only variations in the current of Q2 are seen be the load. So Vout=A*(V1-V2)/2. The load is R1 in parallel with the input impedance of the next stage. The gain depends on the load impedance so the presence of R1 only reduces the gain.
By using a current mirror, Q5 and Q6, you get an active load for Q4 and at the same time restore the gain of the differential stage to Vout=A*(V1-V2). This happens because a variation in the current of Q3 is no longer lost, but gets mirrored via Q5 into Q6, so the excess current flows into/ out of the load, which is actually the input of the next stage. Now Q3 constributes the same amount as Q4 (the changes in the currents of Q3, Q4 are equal and opposite). So Vout=A*(V1-V2). The gain will be dictated only by the input impedance of the next stage, since there is no resistor in parallel with it.
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