In the photo provided the Quantum Fourier Transform is depicted in Qiskit before the barrier. I don't understand the result of inverse. Conceptually, should the inverse of the QFT be the same resultant as the original implementation? And is the inverse correctly depicted in Qiskit for this 3 bit QFT algorithm show below? 
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From linear algebra we know that $(AB)^{-1} = B^{-1} A^{-1}$. This is because $(AB)*(AB)^{-1} = ABB^{-1}A^{-1} = AIA^{-1} = I$.
Hence, if you have the circuit to generate the Bell state from the state $|00\rangle$ which takes the form of:
then its inverse would be:
Putting them together gives you:
This makes sense as the inverse of $CNOT$ is $CNOT$, so they would formed Identity operator. Similarly, the inverse of Hadamard gate is itself... so altogether they will cancel each other out completely.
Thus, if I generate the QFT circuit for 4 qubit without the SWAP at the end, I would have:
then similarly to the above explanation with respect to the bell circuit and its inverse, the inverse QFT would be:
This is the reason why you see that you see in your circuit. hope that helps...
