# 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:

$$\frac{1}{6}\ln(3)-\frac{1}{4}\ln(2)-\frac{1}{12}$$

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What's the answer supposed to be? –  huon-dbaupp Aug 6 '12 at 13:28
If $dy/dx=2/(x+3)^2$, then $1/(x+3)^2=(1/2)dy/dx$. You've used $2dy/dx$ instead. –  celtschk Aug 6 '12 at 13:33
@Auke I thought that homework tag is supposed to be added only by OP (or if the OP states explicitly that it is a homework); see meta. Here's link to revision history. –  Martin Sleziak Aug 6 '12 at 13:45
It's not homework, this is the summer. –  Magpie Aug 6 '12 at 13:49
...and that's why we don't add the homework tag willy-nilly. :| –  Guess who it is. Aug 6 '12 at 13:51

• 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:

$$\dfrac 12 \int_\frac{1}{3}^\frac{1}{2}{\ln(y)dy}=\dfrac 12 \bigg[y(\ln(y)-1\bigg]_\frac{1}{3}^\frac{1}{2} =\\\frac 12 \left(\frac{1}{2}\left(\ln\left(\frac{1}{2}\right)-1\right)-\frac{1}{3}\left(\ln\left(\frac{1}{3}\right)-1\right)\right) \\=\frac 12 \left(-\frac 12 \ln 2 - \frac 12 +\frac 13 \ln 3+ \frac 13\right)\\=\frac 16 \ln 3 -\frac 14 \ln 2 -\frac 1{12}$$

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What do you mean the other issue might be? Once I did the $\frac{1}{2}$ I had the right answer. That what you've written needs correcting I think. –  Magpie Aug 6 '12 at 14:22
@Magpie: Well, I thought you did not notice $\ln \frac 12 = - \ln 2$. –  Gigili Aug 6 '12 at 14:24
Ok, I think I was ok with the logs but just to chekc I have put up what I did further up in the question. LEt me know if you see any problems with it.thanks –  Magpie Aug 6 '12 at 14:45
@Magpie: No, that doesn't seem correct. You have $\int \ln x dx= x(\ln x -1)+C$. –  Gigili Aug 6 '12 at 14:51
ok I have added a step to show what i was thinking, but I have to admit, I don't really get what you mean. –  Magpie Aug 6 '12 at 16:56

The solution I have now is:

$$\frac{1}{2}\int_\frac{1}{3}^\frac{1}{2}\ln(y)dy=\frac{1}{2}\left[y(ln(y)-1)\right]_\frac{1}{3}^\frac{1}{2}$$

$$=\frac{1}{2}\left[\frac{1}{2}\ln(\frac{1}{2})-\frac{1}{2}-\frac{1}{3}\ln(\frac{1}{3})+\frac{1}{3}\right]\\$$

$$=\frac{1}{2}\ln\left(\frac{1}{2}\right)-\frac{1}{4}-\frac{1}{3}\ln\left(\frac{1}{3}\right)+\frac{1}{6}$$

$$=-\frac{1}{4}\ln(2)+\frac{1}{6}\ln({3})-\frac{6}{24}+\frac{4}{24}$$

$$=-\frac{1}{4}\ln(2)+\frac{1}{6}\ln({3})-\frac{1}{12}$$

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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.

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I posted the answer now, the reason I stopped there was that it was not going toward the answer needed. See above. –  Magpie Aug 6 '12 at 13:37
also can you elaborate on why I should multiply but a half instead of 2? Thanks –  Magpie Aug 6 '12 at 13:43
@Magpie: For your second question, see my comment to your post. In short, if $dy/dx=2f$, then $f=\frac12 dy/dx$, not $2dy/dx$. –  celtschk Aug 6 '12 at 13:47
yeah that works. Thanks. –  Magpie Aug 6 '12 at 14:00