integral of (1+3z)(2sqrt(z)-5)

解

\int (1 + 3 z) (2 z - 5) d z

\frac{4}{3} z^{\frac{3}{2}} - 5 z + \frac{12}{5} z^{\frac{5}{2}} - \frac{15 z ^{2}}{2} + C

解答ステップ

\int (1 + 3 z) (2 z - 5) d z

拡張

(1 + 3 z) (2 z - 5) : 2 z - 5 + 6 z^{\frac{3}{2}} - 15 z

= \int 2 z - 5 + 6 z^{\frac{3}{2}} - 15 z 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 2 z d z - \int 5 d z + \int 6 z^{\frac{3}{2}} d z - \int 15 z d z

\int 2 z d z = \frac{4}{3} z^{\frac{3}{2}}

\int 5 d z = 5 z

\int 6 z^{\frac{3}{2}} d z = \frac{12}{5} z^{\frac{5}{2}}

\int 15 z d z = \frac{15 z ^{2}}{2}

= \frac{4}{3} z^{\frac{3}{2}} - 5 z + \frac{12}{5} z^{\frac{5}{2}} - \frac{15 z ^{2}}{2}

定数を解答に追加する

= \frac{4}{3} z^{\frac{3}{2}} - 5 z + \frac{12}{5} z^{\frac{5}{2}} - \frac{15 z ^{2}}{2} + C