And #50 we have Y double prime equaling the quantity X + y prime squared. So here we have our Y prime equaling our DYDX, which is our P. And because there's no single Y but there is a single X, we get our Y double prime equaling P prime or just DPDX. So doing our straight substitution here, our Y prime double prime we could think of as P prime, and that's going to equal X + P quantity squared. Now if I do a substitution, let's let V equal X + P, then our DVDX is just 1 + d PDX or our DVD X - 1 is just DPDX. And that's of course what this P prime is. So we're going to get DVD X - 1 equaling v ^2. So DVD X = v ^2 + 1 DV over v ^2 + 1 equal the integral of DX by taking the integral of each side. This is tangent inverse of V equaling X plus some constant. So X + P is going to equal the tangent of X + C. Now remember that P is really just Y prime, so we get tangent X + C -, X. Now, if we want to integrate here, we're going to integrate our DY, We're going to integrate tangent X + C -, X, all of that being DX. So we're going to get Y on this side. And the tangent of X + C is really the natural log of secant of X + C -1 half X ^2 + a new C.