# How do you use partial fraction decomposition to decompose the fraction to integrate (z+1)/(z^2(z^2+4))?

Jul 4, 2018

$\frac{1}{8} \ln \left({z}^{2} / \left({z}^{2} + 4\right)\right) - \frac{1}{4 z} - \frac{1}{8} {\tan}^{-} 1 \left(\frac{z}{2}\right) + C$

#### Explanation:

While one could use the standard approach of beginning with

$\frac{z + 1}{{z}^{2} \left({z}^{2} + 4\right)} = \frac{A}{z} + \frac{B}{z} ^ 2 + \frac{C z + D}{{z}^{2} + 4}$

it will be faster in this case to note that

$\frac{1}{{z}^{2} \left({z}^{2} + 4\right)} = \frac{1}{4} \left(\frac{1}{z} ^ 2 - \frac{1}{{z}^{2} + 4}\right)$

and thus

$\frac{z + 1}{{z}^{2} \left({z}^{2} + 4\right)} = \frac{z + 1}{4} \left(\frac{1}{z} ^ 2 - \frac{1}{{z}^{2} + 4}\right)$

$q \quad q \quad = \frac{1}{4} \left(\frac{z + 1}{z} ^ 2 - \frac{z + 1}{{z}^{2} + 4}\right)$

$q \quad q \quad = \frac{1}{4} \left(\frac{1}{z} + \frac{1}{z} ^ 2 - \frac{z + 1}{{z}^{2} + 4}\right)$

Thus

$\int \frac{z + 1}{{z}^{2} \left({z}^{2} + 4\right)} \mathrm{dz} = \int \frac{1}{4} \left(\frac{1}{z} + \frac{1}{z} ^ 2 - \frac{z + 1}{{z}^{2} + 4}\right) \mathrm{dz}$

$q \quad = \frac{1}{4} \log z - \frac{1}{4 z} - \frac{1}{8} \ln \left({z}^{2} + 4\right) - \frac{1}{8} {\tan}^{-} 1 \left(\frac{z}{2}\right) + C$

$q \quad = \frac{1}{8} \ln \left({z}^{2} / \left({z}^{2} + 4\right)\right) - \frac{1}{4 z} - \frac{1}{8} {\tan}^{-} 1 \left(\frac{z}{2}\right) + C$