integration 0 to infinity for dx/(x+2)(2x+1)

[pixelmath]

Hi All, I am a first timer here

could someone explain me through the following problem, please
integration 0 to infinity for dx/(x+2)(2x+1)
I am looking for steps for understanding, not the direct answer

 

[MarkFL]

I would first use partial fraction decomposition on the integrand...what do you get for the integrand when you do that?

 

[pixelmath]

stuck at x->inf ( ln((2x+1)/(x+2))

 

[MarkFL]

As I mentioned, I would write:

[MATH]\frac{1}{(x+2)(2x+1)}=\frac{1}{3}\left(\frac{2}{2x+1}-\frac{1}{x+2}\right)[/MATH]
And so our improper integral can not be expressed as:

[MATH]I=\frac{1}{3}\lim_{t\to\infty}\left(\int_0^t \frac{2}{2x+1}-\frac{1}{x+2}\,dx\right)[/MATH]
Using the FTOC, this becomes:

[MATH]I=\frac{1}{3}\lim_{t\to\infty}\left(\left[\ln\left(\frac{2x+1}{x+2}\right)\right]_0^t\right)[/MATH]
[MATH]I=\frac{1}{3}\lim_{t\to\infty}\left(\ln\left(\frac{2t+1}{t+2}\right)-\ln\left(\frac{1}{2}\right)\right)[/MATH]
Is this essentially where you're stuck?

 

[pixelmath]

yes exactly, also how were you able to identify

is there any easy way to identify or was it through intuition ?

 

[MarkFL]

There are shortcut methods, but this is the way I was taught to do partial fractions:

Assume that the given expression has the following decomposition:

[MATH]\frac{1}{(x+2)(2x+1)}=\frac{A}{x+2}+\frac{B}{2x+1}[/MATH]
Multiply by \((x+2)(2x+1)\) to get:

[MATH]1=A(2x+1)+B(x+2)[/MATH]
Distribute and arrange as:

[MATH]0x+1=(2A+B)x+(A+2B)[/MATH]
Equating corresponding coefficients, we obtain the system:

[MATH]2A+B=0[/MATH]
[MATH]A+2B=1[/MATH]
Solving this system we obtain:

[MATH](A,B)=\left(-\frac{1}{3},\frac{2}{3}\right)[/MATH]
And so we know:

[MATH]\frac{1}{(x+2)(2x+1)}=\frac{-\dfrac{1}{3}}{x+2}+\frac{\dfrac{2}{3}}{2x+1}=\frac{1}{3}\left(\frac{2}{2x+1}-\frac{1}{x+2}\right)[/MATH]
Back to the integral, suppose we write it as:

[MATH]I=\frac{1}{3}\lim_{t\to\infty}\left(\ln\left(\frac{2t+1}{t+2}\right)+\ln(2)\right)[/MATH]
[MATH]I=\frac{1}{3}\lim_{t\to\infty}\left(\ln\left(\frac{2(2t+1)}{t+2}\right)\right)[/MATH]
[MATH]I=\frac{1}{3}\lim_{t\to\infty}\left(\ln\left(\frac{2\left(2+\dfrac{1}{t}\right)}{1+\dfrac{2}{t}}\right)\right)[/MATH]
Can you proceed?