integral of ((z+2))/(z^2+3z-1)

解

\int \frac{( z + 2 )}{z ^{2} + 3 z - 1} d z

\frac{1}{2} ln 4 z^{2} + 12 z - 4 - \frac{1}{2 13} ln \frac{1}{13} (2 z + 3) + 1 + \frac{1}{2 13} ln \frac{1}{13} (2 z + 3) - 1 + C

解答ステップ

\int \frac{z + 2}{z ^{2} + 3 z - 1} d z

拡張

\frac{z + 2}{z ^{2} + 3 z - 1} : \frac{z}{z ^{2} + 3 z - 1} + \frac{2}{z ^{2} + 3 z - 1}

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 ^{2} + 3 z - 1} d z + \int \frac{2}{z ^{2} + 3 z - 1} d z

\int \frac{z}{z ^{2} + 3 z - 1} d z = 2 (\frac{1}{4} ln 4 z^{2} + 12 z - 4 + \frac{3}{4 13} (ln \frac{2 z + 3}{13} + 1 - ln \frac{2 z + 3}{13} - 1))

\int \frac{2}{z ^{2} + 3 z - 1} d z = - \frac{2}{13} (ln \frac{2 z + 3}{13} + 1 - ln \frac{2 z + 3}{13} - 1)

= 2 (\frac{1}{4} ln 4 z^{2} + 12 z - 4 + \frac{3}{4 13} (ln \frac{2 z + 3}{13} + 1 - ln \frac{2 z + 3}{13} - 1)) - \frac{2}{13} (ln \frac{2 z + 3}{13} + 1 - ln \frac{2 z + 3}{13} - 1)

簡素化

2 (\frac{1}{4} ln 4 z^{2} + 12 z - 4 + \frac{3}{4 13} (ln \frac{2 z + 3}{13} + 1 - ln \frac{2 z + 3}{13} - 1)) - \frac{2}{13} (ln \frac{2 z + 3}{13} + 1 - ln \frac{2 z + 3}{13} - 1) : \frac{1}{2} ln 4 z^{2} + 12 z - 4 - \frac{1}{2 13} ln \frac{1}{13} (2 z + 3) + 1 + \frac{1}{2 13} ln \frac{1}{13} (2 z + 3) - 1

= \frac{1}{2} ln 4 z^{2} + 12 z - 4 - \frac{1}{2 13} ln \frac{1}{13} (2 z + 3) + 1 + \frac{1}{2 13} ln \frac{1}{13} (2 z + 3) - 1

定数を解答に追加する

= \frac{1}{2} ln 4 z^{2} + 12 z - 4 - \frac{1}{2 13} ln \frac{1}{13} (2 z + 3) + 1 + \frac{1}{2 13} ln \frac{1}{13} (2 z + 3) - 1 + C