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人気のある微分積分 >

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

解

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

解

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

+1

十進法表記

0.09776 \dots

解答ステップ

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

定数を除く:

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

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

以下の部分分数を得る:

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

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

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

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

\int_{2}^{3} \frac{1}{x ^{2}} d x = \frac{1}{6}

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

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

拡張

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

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

グラフ

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