integral from 1 to 2 of (x^2-x-5)/(x+2)

解

\int_{1}^{2} \frac{x ^{2} - x - 5}{x + 2} d x

2 ln (2) - ln (3) - \frac{3}{2}

十進法表記

- 1.21231 \dots

解答ステップ

\int_{1}^{2} \frac{x ^{2} - x - 5}{x + 2} d x

拡張

\frac{x ^{2} - x - 5}{x + 2} : \frac{x ^{2}}{x + 2} - \frac{x}{x + 2} - \frac{5}{x + 2}

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_{1}^{2} \frac{x ^{2}}{x + 2} d x - \int_{1}^{2} \frac{x}{x + 2} d x - \int_{1}^{2} \frac{5}{x + 2} d x

\int_{1}^{2} \frac{x ^{2}}{x + 2} d x = 4 (2 ln (2) - ln (3)) - \frac{1}{2}

\int_{1}^{2} \frac{x}{x + 2} d x = 1 - 2 (2 ln (2) - ln (3))

\int_{1}^{2} \frac{5}{x + 2} d x = 5 (2 ln (2) - ln (3))

= 4 (2 ln (2) - ln (3)) - \frac{1}{2} - (1 - 2 (2 ln (2) - ln (3))) - 5 (2 ln (2) - ln (3))

簡素化

4 (2 ln (2) - ln (3)) - \frac{1}{2} - (1 - 2 (2 ln (2) - ln (3))) - 5 (2 ln (2) - ln (3)) : 2 ln (2) - ln (3) - \frac{3}{2}

= 2 ln (2) - ln (3) - \frac{3}{2}