Is the Integration Calculation Correct?

[suspenc3]

\int_0^* \frac {dz}{z^23Z +2}

=\lim_{t\rightarrow \*} \int_0^* \frac{dz}{z^2+3Z+2}

=\lim_{t\rightarrow \*} \int_0^* \frac{dz}{(z+3/2)^2 -1/4}

let u=z+3/2
du=dz

=\lim_{t\rightarrow \*} \int_0^* \frac{du}{(u)^2 -1/4}

= \frac{1}{2(1/2)}ln|\frac{u-1/2}{u+1/2}

= ln |\frac{z+1}{z+2}| \right]_0^*

= -ln \frac{1}{2}

this is one of the first of these I am doing so bear with me if its horrible wrong

EDIT:* denotes infinite..PS is there a latex for infinite?

 

[Gagle The Terrible]

I don't know why you used the lim symbol in the first place, and where the t commes in play, but

= \int_0^* \frac{dz}{z^2+3Z+2}
= \int_0^* \frac{du}{(u)^2 -1/4}

using your algebraic manipulations. After that, you used the partial sum decomposition skillfully and unless I'm mistaking, you are right!

Just one minor thing : -ln \frac{1}{2} = ln(2)

 

[suspenc3]

yyyea..but what?I already had that is it right?

 

[0rthodontist]

It's correct except you forgot to change the lower limit of integration. When z goes from 0 to infinity, then u = z + 3/2 goes from what to what?

You do the infinity sign like \infty