If all of that sounds really confusing, I'll actually do the u substitution. So if I were just trying to evaluate, if I were just trying to evaluate the integral two over x minus one dx, I could see, okay, the derivative of x minus one is just one. So I could say u is equal to x minus one. And then du is going to be equal to dx. And so this is going to be, we can rewrite in terms of u as two, I'll just take the constant out, two times the integral of one over u du, which we know is two times the natural log of the absolute value of u plus c. And in this case, we know that u is x minus one. So this is equal to two times the natural log of x minus one plus c. And so that's what we're gonna have right over here. So plus two times the natural log of the absolute value of x minus one plus c. And the plus c doesn't just come from this one.