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人気のある微分積分 >

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

解

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

解

- ln (2) + 2

+1

十進法表記

1.30685 \dots

解答ステップ

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

u置換積分法を適用する

= \int_{2}^{4} \frac{u ^{2} - 2 u - 1}{u} d u

拡張

\frac{u ^{2} - 2 u - 1}{u} : u - 2 - \frac{1}{u}

= \int_{2}^{4} u - 2 - \frac{1}{u} 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

= \int_{2}^{4} u d u - \int_{2}^{4} 2 d u - \int_{2}^{4} \frac{1}{u} d u

\int_{2}^{4} u d u = 6

\int_{2}^{4} 2 d u = 4

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

= 6 - 4 - ln (2)

簡素化

= - ln (2) + 2

グラフ

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