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人気のある微分積分 >

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

解

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

解

3 + ln (2)

+1

十進法表記

3.69314 \dots

解答ステップ

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

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

拡張

\frac{2 x + 1}{x} : 2 + \frac{1}{x}

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

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

\int_{1}^{4} 2 d x = 6

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

= \frac{1}{2} (6 + 2 ln (2))

簡素化

\frac{1}{2} (6 + 2 ln (2)) : 3 + ln (2)

= 3 + ln (2)

グラフ

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