Evaluating $\int_0^1{\frac{1}{(x+3)^2}}\ln\left(\frac{x+1}{x+3}\right)dx$ using $\frac{dy}{dx}=\frac{2}{(x+3)^2}$ where $y=\frac{x+1}{x+3}$ 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:
$$\frac{1}{6}\ln(3)-\frac{1}{4}\ln(2)-\frac{1}{12}$$
 A: *

*You have:


$$\dfrac{dy}{dx}=\dfrac{(x+3)}{(x+3)^2}-\dfrac{(x+1)}{(x+3)^2}=\dfrac{2}{(x+3)^2}$$
But the integral is $I=\int_0^1{\dfrac{1}{(x+3)^2}}\ln\left(\dfrac{x+1}{x+3}\right)dx$ where ${\dfrac{1}{(x+3)^2}}$ is actually $\dfrac 12 \times \dfrac{2}{(x+3)^2}$. Therefore:
$$I= \dfrac 12 \int_0^1{\frac{dy}{dx}}\ln(y)dx=\dfrac 12 \int_\frac{1}{3}^\frac{1}{2}{\ln(y)dy}$$


*

*The other issue might be:


$$\dfrac 12 \int_\frac{1}{3}^\frac{1}{2}{\ln(y)dy}=\dfrac 12 \bigg[y(\ln(y)-1\bigg]_\frac{1}{3}^\frac{1}{2}
=\\\frac 12 \left(\frac{1}{2}\left(\ln\left(\frac{1}{2}\right)-1\right)-\frac{1}{3}\left(\ln\left(\frac{1}{3}\right)-1\right)\right)
\\=\frac 12 \left(-\frac 12 \ln 2 - \frac 12 +\frac 13 \ln 3+ \frac 13\right)\\=\frac 16 \ln 3 -\frac 14 \ln 2 -\frac 1{12}$$
A: The solution I have now is:
$$\frac{1}{2}\int_\frac{1}{3}^\frac{1}{2}\ln(y)dy=\frac{1}{2}\left[y(ln(y)-1)\right]_\frac{1}{3}^\frac{1}{2}$$
$$=\frac{1}{2}\left[\frac{1}{2}\ln(\frac{1}{2})-\frac{1}{2}-\frac{1}{3}\ln(\frac{1}{3})+\frac{1}{3}\right]\\$$
$$=\frac{1}{2}\ln\left(\frac{1}{2}\right)-\frac{1}{4}-\frac{1}{3}\ln\left(\frac{1}{3}\right)+\frac{1}{6}$$
$$=-\frac{1}{4}\ln(2)+\frac{1}{6}\ln({3})-\frac{6}{24}+\frac{4}{24}$$
$$=-\frac{1}{4}\ln(2)+\frac{1}{6}\ln({3})-\frac{1}{12}$$
A: You should have been careful when multiplying by $2$, I think that you should have multiplied by $\frac{1}{2}$ in your formal calculation.
Also, you may need to rewrite your answer a bit to get it right, such as computing $\frac{2}{9} - 1 = -\frac{7}{9}$, which is implicitly required from you by any textbook, I presume.
