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人気のある微分積分 >

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

解

\int_{2}^{3} \frac{2 x ^{2}}{x ^{2} - 1} d x

解

- ln (2) + ln (3) + 2

+1

十進法表記

2.40546 \dots

解答ステップ

\int_{2}^{3} \frac{2 x ^{2}}{x ^{2} - 1} 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{x ^{2}}{x ^{2} - 1} d x

因数

x^{2} - 1 : - (- x^{2} + 1)

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

定数を除く:

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

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

\frac{x ^{2}}{- x ^{2} + 1} = \frac{1}{- x ^{2} + 1} - 1

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

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

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

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

簡素化

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

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

グラフ

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