What is the integral of (z+5)/(z(z^2+4)) ?

The integral of (z+5)/(z(z^2+4)) is 1/2 arctan(z/2)+5(1/4 ln|z|-1/8 ln|z^2+4|)+C

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integral of (z+5)/(z(z^2+4))

\int \frac{z + 5}{z ( z ^{2} + 4 )} d z

\frac{1}{2} arctan (\frac{z}{2}) + 5 (\frac{1}{4} ln ∣ z ∣ - \frac{1}{8} ln z^{2} + 4) + C

Solution steps

\int \frac{z + 5}{z ( z ^{2} + 4 )} d z

Expand

\frac{z + 5}{z ( z ^{2} + 4 )} : \frac{z}{z ( z ^{2} + 4 )} + \frac{5}{z ( z ^{2} + 4 )}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{z}{z ( z ^{2} + 4 )} d z + \int \frac{5}{z ( z ^{2} + 4 )} d z

\int \frac{z}{z ( z ^{2} + 4 )} d z = \frac{1}{2} arctan (\frac{z}{2})

\int \frac{5}{z ( z ^{2} + 4 )} d z = 5 (\frac{1}{4} ln ∣ z ∣ - \frac{1}{8} ln z^{2} + 4)

= \frac{1}{2} arctan (\frac{z}{2}) + 5 (\frac{1}{4} ln ∣ z ∣ - \frac{1}{8} ln z^{2} + 4)

Add a constant to the solution

= \frac{1}{2} arctan (\frac{z}{2}) + 5 (\frac{1}{4} ln ∣ z ∣ - \frac{1}{8} ln z^{2} + 4) + C

\frac{\partial}{\partial v} (u v^{2} + w^{3})

\int e^{5 x + 6} d x

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{2 t}

f (x) = x - 2 sin (x), at x = 0

x y^{^{''}} - y^{^{'}} = x^{2}

What is the integral of (z+5)/(z(z^2+4)) ?
The integral of (z+5)/(z(z^2+4)) is 1/2 arctan(z/2)+5(1/4 ln|z|-1/8 ln|z^2+4|)+C