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人気のある微分積分 >

integral from 0 to 2 of (x^3)/(x^2+1)

解

\int_{0}^{2} \frac{x ^{3}}{x ^{2} + 1} d x

解

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

+1

十進法表記

1.19528 \dots

解答ステップ

\int_{0}^{2} \frac{x ^{3}}{x ^{2} + 1} d x

u置換積分法を適用する

= \int_{1}^{5} \frac{u - 1}{2 u} 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_{1}^{5} \frac{u - 1}{u} d u

拡張

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

= \frac{1}{2} \cdot \int_{1}^{5} 1 - \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

= \frac{1}{2} (\int_{1}^{5} 1 d u - \int_{1}^{5} \frac{1}{u} d u)

\int_{1}^{5} 1 d u = 4

\int_{1}^{5} \frac{1}{u} d u = ln (5)

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

グラフ

人気の例

(\partial)/(\partial x)((yz)/(x^2+y^2))

\frac{\partial}{\partial x} (\frac{yz}{x ^{2} + y ^{2}})

derivative f(x)=-9e^{x^{-9}}

d er i v a t i v e f (x) = - 9 e^{x^{- 9}}

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derivative of 1/(\sqrt[5]{x})

\frac{d}{d x} (\frac{1}{5 x})

limit as x approaches-2 of 3x^2+4x-2

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