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人気のある微分積分 >

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

解

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

解

\frac{112 ln ( 2 ) - 28 ln ( 7 ) - 9}{504}

+1

十進法表記

0.02806 \dots

解答ステップ

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

u置換積分法を適用する

= \int_{4}^{7} \frac{1}{2 ( u - 3 ) u ^{2}} d u

定数を除く:

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

= \frac{1}{2} \cdot \int_{4}^{7} \frac{1}{( u - 3 ) u ^{2}} d u

以下の部分分数を得る:

\frac{1}{( u - 3 ) u ^{2}} : \frac{1}{9 ( u - 3 )} - \frac{1}{9 u} - \frac{1}{3 u ^{2}}

= \frac{1}{2} \cdot \int_{4}^{7} \frac{1}{9 ( u - 3 )} - \frac{1}{9 u} - \frac{1}{3 u ^{2}} d u

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_{4}^{7} \frac{1}{9 ( u - 3 )} d u - \int_{4}^{7} \frac{1}{9 u} d u - \int_{4}^{7} \frac{1}{3 u ^{2}} d u)

\int_{4}^{7} \frac{1}{9 ( u - 3 )} d u = \frac{2}{9} ln (2)

\int_{4}^{7} \frac{1}{9 u} d u = \frac{1}{9} (ln (7) - 2 ln (2))

\int_{4}^{7} \frac{1}{3 u ^{2}} d u = \frac{1}{28}

= \frac{1}{2} (\frac{2}{9} ln (2) - \frac{1}{9} (ln (7) - 2 ln (2)) - \frac{1}{28})

簡素化

\frac{1}{2} (\frac{2}{9} ln (2) - \frac{1}{9} (ln (7) - 2 ln (2)) - \frac{1}{28}) : \frac{112 ln ( 2 ) - 28 ln ( 7 ) - 9}{504}

= \frac{112 ln ( 2 ) - 28 ln ( 7 ) - 9}{504}

グラフ

人気の例

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