Question

Evaluate the ∫(z+2)/(z²+4z) dz from -3 to 2

Answer 1

$$\displaystyle\int_{-3}^2 \dfrac{z+2}{z^2+4z}dz$$

$$\int \dfrac{z+2}{z^2+4z}dz$$

Use $u$-substitution

$u=z^2+4z, du=(2z+4)dz$

$$\int \dfrac{z+2}{z^2+4z}dz=\dfrac{1}{2}\int \dfrac{du}{u}=\dfrac{1}{2}\ln|u|+C$$

$$=\dfrac{1}{2}\ln|z^2+4z|+C$$

$$\displaystyle\int_{-3}^2 \dfrac{z+2}{z^2+4z}dz=\displaystyle\int_{-3}^0 \dfrac{z+2}{z^2+4z}dz+\displaystyle\int_{0}^2 \dfrac{z+2}{z^2+4z}dz$$

$$=\lim\limits_{t\to0^-}\displaystyle\int_{-3}^t \dfrac{z+2}{z^2+4z}dz+\lim\limits_{t\to0^+}\displaystyle\int_{t}^2 \dfrac{z+2}{z^2+4z}dz$$

$$=\lim\limits_{t\to0^-}\bigg[\dfrac{1}{2}\ln|z^2+4z|\bigg]\begin{matrix} t \\ -3 \end{matrix}+\lim\limits_{t\to0^+}\bigg[\dfrac{1}{2}\ln|z^2+4z|\bigg]\begin{matrix} 2 \\ t \end{matrix}$$

$$=\dfrac{1}{2}\lim\limits_{t\to0^-}\ln|t^2+4t|-\dfrac{1}{2}\lim\limits_{t\to0^-}\ln|(-3)^2+4(-3)|+$$

$$+\dfrac{1}{2}\ln|(2)^2+4(2)|-\dfrac{1}{2}\lim\limits_{t\to0^+}\ln|t^2+4t|$$

The integral $\displaystyle\int_{-3}^2 \dfrac{z+2}{z^2+4z}dz$ diverges.