# ∫▒(2x+1)/((x+2)^2) dx A problem with Partial fractions and integration[Attempt to sv]

• Jan 21st 2011, 01:53 AM
Riazy
∫▒(2x+1)/((x+2)^2) dx A problem with Partial fractions and integration[Attempt to sv]
Hello guys, here is the problem below:

Attachment 20524
• Jan 21st 2011, 02:01 AM
TheCoffeeMachine
Because you have a case of multiple root in the denominator,
the correct partial fraction decomposition set-up should be:

$\displaystyle \frac{2x+1}{(x+2)^2} = \frac{A}{x+2}+\frac{B}{(x+2)^2}$.
• Jan 21st 2011, 02:04 AM
FernandoRevilla
$\dfrac{2x+1}{(x+2)^2}=\dfrac{A}{x+2}+\dfrac{B}{(x+ 2)^2}=\dfrac{A(x+2)+B}{(x+2)^2}$

$A(x+2)+B=2x+1\Rightarrow A=2,B=-3$

Fernando Revilla

Edited: Sorry, I didn't see TheCoffeeMachine's post.
• Jan 21st 2011, 02:14 AM
Riazy
I didn't get how you got to this through getting the same denominator: \dfrac{2x+1}{(x+2)^2}=\dfrac{A}{x+2}+\dfrac{B}{(x+ 2)^2}=\dfrac{A(x+2+B}{(x+2)^2} ( The last part)

Because i get A (x+2)^2 + B(x+2) / (x+2)(x+2)^2
• Jan 21st 2011, 02:18 AM
Prove It
An alternative:

$\displaystyle \int{\frac{2x + 1}{(x + 2)^2}\,dx} = \int{\frac{2x + 4 - 3}{(x + 2)^2}\,dx}$

$\displaystyle = \int{\frac{2x + 4}{x^2 + 4x + 4}\,dx} - 3\int{(x + 2)^{-2}\,dx}$.

Substitute $\displaystyle u = x^2 + 4x + 4$ so that $\displaystyle du = (2x + 4)\,dx$ and substitute $\displaystyle v = x + 2$ so that $\displaystyle dv = dx$ and the integrals become

$\displaystyle \int{\frac{1}{u}\,du} - 3\int{v^{-2}\,dv}$.

I'm sure you can go from here.
• Jan 21st 2011, 02:19 AM
Riazy
;) thanks
• Jan 21st 2011, 02:27 AM
FernandoRevilla
Quote:

Originally Posted by Riazy
I didn't get how you got to this through getting the same denominator: \dfrac{2x+1}{(x+2)^2}=\dfrac{A}{x+2}+\dfrac{B}{(x+ 2)^2}=\dfrac{A(x+2+B}{(x+2)^2} ( The last part)

Because i get A (x+2)^2 + B(x+2) / (x+2)(x+2)^2

Divide numerator and denominator by $x+2$ .

Fernando Revilla