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人気のある微分積分 >

integral from 1 to 2 of 4/(x^3+x)

解

\int_{1}^{2} \frac{4}{x ^{3} + x} d x

解

2 (3 ln (2) - ln (5))

+1

十進法表記

0.94000 \dots

解答ステップ

\int_{1}^{2} \frac{4}{x ^{3} + x} d x

定数を除く:

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

= 4 \cdot \int_{1}^{2} \frac{1}{x ^{3} + x} d x

以下の部分分数を得る:

\frac{1}{x ^{3} + x} : \frac{1}{x} - \frac{x}{x ^{2} + 1}

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

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

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

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

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

簡素化

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

= 2 (3 ln (2) - ln (5))

グラフ

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