Find derivative of $y= \frac{ax+b}{cx+d}$
I found it to be $\frac{dy}{dx}=\frac{a}{cx+d}-\frac{c(ax+b)}{(cx+d)^2}$
Use it to evaluate:
$\int_0^1{\frac{1}{(x+3)^2}}\ln\left(\frac{x+1}{x+3}\right)dx$
I figured that here $y=\frac{x+1}{x+3}$ and $\frac{dy}{dx}=\frac{1}{x+3}-\frac{(x+1)}{(x+3)^2}$
and using the technique I learned from my last question I did this:
$\frac{dy}{dx}=\frac{(x+3)}{(x+3)^2}-\frac{(x+1)}{(x+3)^2}=\frac{2}{(x+3)^2}$
which I could then substitute back, having changed the limits by substituting $1$ into $y$ and then $0$ into $y$:
$y|_{x=1}=\frac{x+1}{x+3}=\frac{1}{2}$
$y|_{x=0}=\frac{1}{3}=\frac{1}{3}$
$2\int_0^1{\frac{dy}{dx}}\ln(y)dx=2\int_\frac{1}{3}^\frac{1}{2}{\ln(y)dy}$
This gives me:
$2\int_\frac{1}{3}^\frac{1}{2}{\ln(y)dy}$
$=2\left[y(\ln(y)-1)\right]_\frac{1}{3}^\frac{1}{2} = 2\left[\frac{1}{2}\left(\ln\left(\frac{1}{2}\right)-\frac{1}{2}\right)\right]-\frac{1}{3}\left[\ln\left(\frac{1}{3}\right)-\frac{1}{3}\right]\\$
$\ln\left(\frac{1}{2}\right)-1-\frac{2}{3}\ln\left(\frac{1}{3}\right)+\frac{2}{9}$
The problem is I am supposed to end up with something else. Can anyone spot any issues with this?
EDIT: This is the answer I am supposed to be getting:
$\frac{1}{6}\ln(3)-\frac{1}{4}\ln(2)-\frac{1}{12}$