Improper Integrals

[thatguy47]

I'm confused with this problem:

Wouldn't the first step be adding them not subtracting?
I don't understand the method their using to solve this problem? Can someone please explain it to me? Thanks.

 

[skeeter]

are you familiar with the method of partial fractions?

$\displaystyle \frac{1}{z^2+3z+2} = \frac{1}{(z+1)(z+2)} = \frac{A}{z+1} + \frac{B}{z+2}$

$\displaystyle \frac{1}{(z+1)(z+2)} = \frac{A(z+2)}{(z+1)(z+2)} + \frac{B(z+1)}{(z+1)(z+2)}$

numerators form the equation ...

$\displaystyle 1 = A(z+2) + B(z+1)$

let $\displaystyle z = -2$ ...

$\displaystyle 1 = B(-1)$ ...$\displaystyle B = -1$

let $\displaystyle z = -1$ ...

$\displaystyle 1 = A(1)$ ... $\displaystyle A = 1$

so,

$\displaystyle \frac{1}{z^2+3z+2} = \frac{1}{z+1} - \frac{1}{z+2}$

 

[royhaas]

What part do you not understand?

 

[thatguy47]

Thanks skeeter.
One more thing, I assume the ln(t+1)/(t+2) at the end equals zero but how does the -ln(1/2) equal ln1 + ln 2 ?
I thought it would be -ln1 + ln2.....I guess it doesn't matter since ln1 = 0.

 

[skeeter]

when they take the limit, the expression becomes

$\displaystyle \ln(1) - \ln\left(\frac{1}{2}\right) = 0 - \ln\left(\frac{1}{2}\right) = 0 + \ln(2)$

remember that

$\displaystyle -\ln\left(\frac{1}{2}\right) = \ln\left(\frac{1}{2}\right)^{-1} = \ln(2)$