integral from 2 to 4 of x/((x+1)(x+2))

解

\int_{2}^{4} \frac{x}{( x + 1 ) ( x + 2 )} d x

- ln (5) + ln (3) + 2 ln (\frac{3}{2})

十進法表記

0.30010 \dots

解答ステップ

\int_{2}^{4} \frac{x}{( x + 1 ) ( x + 2 )} d x

以下の部分分数を得る:

\frac{x}{( x + 1 ) ( x + 2 )} : - \frac{1}{x + 1} + \frac{2}{x + 2}

= \int_{2}^{4} - \frac{1}{x + 1} + \frac{2}{x + 2} d x

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}^{4} \frac{1}{x + 1} d x + \int_{2}^{4} \frac{2}{x + 2} d x

\int_{2}^{4} \frac{1}{x + 1} d x = ln (5) - ln (3)

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

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

簡素化

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