ddx(xx+1)\frac{d}{dx}\left(\frac{x}{x+1}\right)dxd(x+1x)) = vdudx−udvdxv2\frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}v2vdxdu−udxdv
u = x, dudx\frac{du}{dx}dxdu = 1, v = x + 1, dvdx\frac{dv}{dx}dxdv = 1
= (x+1)(1)−x(1)(x+1)2\frac{(x+1)(1)-x(1)}{(x+1)^2}(x+1)2(x+1)(1)−x(1)
= x+1−x(x+1)2\frac{x+1-x}{(x+1)^2}(x+1)2x+1−x
= 1(x+1)2\frac{1}{(x+1)^2}(x+1)21
