1
$\begingroup$

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:

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

  • 2
    If $dy/dx=2/(x+3)^2$, then $1/(x+3)^2=(1/2)dy/dx$. You've used $2dy/dx$ instead.2012-08-06

3 Answers 3

0

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.

  • 0
    yeah that works. Thanks.2012-08-06
3
  • 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:

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

  • 0
    @Gigali. I still think it should be $-\frac{1}{12}$. Check that please.2012-08-06
1

The solution I have now is:

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

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

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

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

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