# Does this look right?

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

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

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

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 im doing so bear with me if its horrible wrong

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

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## Answers and Replies

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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)

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

0rthodontist
You do the infinity sign like $$\infty$$