integral of (6-z^2)/(2z^2+z-21)

解

\int \frac{6 - z ^{2}}{2 z ^{2} + z - 21} d z

- \frac{1}{2} z + \frac{1}{26} (26 (\frac{1}{8} ln 16 z^{2} + 8 z - 168 + \frac{ln \frac{14}{13} + \frac{4}{13} z - ln - \frac{12}{13} + \frac{4}{13} z}{104}) + 9 (ln \frac{4}{13} z + \frac{14}{13} - ln \frac{4}{13} z - \frac{12}{13})) + C

解答ステップ

\int \frac{6 - z ^{2}}{2 z ^{2} + z - 21} d z

長除法

\frac{6 - z ^{2}}{2 z ^{2} + z - 21} : - \frac{1}{2} + \frac{z - 9}{2 ( 2 z ^{2} + z - 21 )}

= \int - \frac{1}{2} + \frac{z - 9}{2 ( 2 z ^{2} + z - 21 )} d z

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{1}{2} d z + \int \frac{z - 9}{2 ( 2 z ^{2} + z - 21 )} d z

\int \frac{1}{2} d z = \frac{1}{2} z

\int \frac{z - 9}{2 ( 2 z ^{2} + z - 21 )} d z = \frac{1}{26} (26 (\frac{1}{8} ln 16 z^{2} + 8 z - 168 + \frac{ln \frac{14}{13} + \frac{4}{13} z - ln - \frac{12}{13} + \frac{4}{13} z}{104}) + 9 (ln \frac{4}{13} z + \frac{14}{13} - ln \frac{4}{13} z - \frac{12}{13}))

= - \frac{1}{2} z + \frac{1}{26} (26 (\frac{1}{8} ln 16 z^{2} + 8 z - 168 + \frac{ln \frac{14}{13} + \frac{4}{13} z - ln - \frac{12}{13} + \frac{4}{13} z}{104}) + 9 (ln \frac{4}{13} z + \frac{14}{13} - ln \frac{4}{13} z - \frac{12}{13}))

定数を解答に追加する

= - \frac{1}{2} z + \frac{1}{26} (26 (\frac{1}{8} ln 16 z^{2} + 8 z - 168 + \frac{ln \frac{14}{13} + \frac{4}{13} z - ln - \frac{12}{13} + \frac{4}{13} z}{104}) + 9 (ln \frac{4}{13} z + \frac{14}{13} - ln \frac{4}{13} z - \frac{12}{13})) + C