integral from 0 to e of ((x^2-9)/(x+3))

解

\int_{0}^{e} (\frac{x ^{2} - 9}{x + 3}) d x

\frac{e ^{2} + 6 e}{2} - 6 e

十進法表記

- 4.46031 \dots

解答ステップ

\int_{0}^{e} \frac{x ^{2} - 9}{x + 3} d x

拡張

\frac{x ^{2} - 9}{x + 3} : \frac{x ^{2}}{x + 3} - \frac{9}{x + 3}

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_{0}^{e} \frac{x ^{2}}{x + 3} d x - \int_{0}^{e} \frac{9}{x + 3} d x

\int_{0}^{e} \frac{x ^{2}}{x + 3} d x = \frac{e ^{2} + 6 e}{2} - 6 e + 9 (ln (e + 3) - ln (3))

\int_{0}^{e} \frac{9}{x + 3} d x = 9 (ln (e + 3) - ln (3))

= \frac{e ^{2} + 6 e}{2} - 6 e + 9 (ln (e + 3) - ln (3)) - 9 (ln (e + 3) - ln (3))

簡素化

= \frac{e ^{2} + 6 e}{2} - 6 e