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

解

\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

解答ステップ

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

拡張

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

総和規則を適用する:

\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)

定数を解答に追加する

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