Hello! Each of the X plus E is equal to the negative X. The quantity E is divided by X minus E. We need to integrate both sides with respect to X. Is it possible that we will be equal to the integral of E. to the X minus E. To the two that are not positive. Then times E to the negative X. A X. It's okay. If we assume that F of X is equal to E to the X plus E to the negative acts, then that's how it will be. The F prime of X is going to be equal to the negative acts of X and E. The half of an accident that is negative two times F prime is going to be equal to the other half, divided by negative one plus C. It is going to be equal to a one over E. The P here is equal to one over E to the X. It could open C minus one over each of the X, not plus but one over E to the X minus E, to the negative X. It is a constant here, we have it. Okay, take care. Yeah, definitely.