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人気のある微分積分 >

integral from 0 to 1 of 1/(3+e^{-2x)}

解

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

解

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

+1

十進法表記

0.29274 \dots

解答ステップ

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

u置換積分法を適用する

= \int_{0}^{- 2} - \frac{1}{2 ( 3 + e ^{u} )} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 2}^{0} - \frac{1}{2 ( 3 + e ^{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_{- 2}^{0} \frac{1}{3 + e ^{u}} d u)

u置換積分法を適用する

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

以下の部分分数を得る:

\frac{1}{v ( 3 + v )} : \frac{1}{3 v} - \frac{1}{3 ( v + 3 )}

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

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_{\frac{1}{e ^{2}}}^{1} \frac{1}{3 v} d v - \int_{\frac{1}{e ^{2}}}^{1} \frac{1}{3 ( v + 3 )} d v))

\int_{\frac{1}{e ^{2}}}^{1} \frac{1}{3 v} d v = \frac{2}{3}

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

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

簡素化

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

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

グラフ

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