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

解

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

ln (2)

十進法表記

0.69314 \dots

解答ステップ

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

以下の部分分数を得る:

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

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

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

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

= 2 ln (2) - ln (2)

簡素化

= ln (2)