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

解

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

1 - ln (2 e + 1) + ln (3)

十進法表記

0.23661 \dots

解答ステップ

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

以下の部分分数を得る:

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

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

\int_{1}^{e} \frac{1}{x} d x = 1

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

= 1 - (ln (2 e + 1) - ln (3))

簡素化

= 1 - ln (2 e + 1) + ln (3)