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人気のある微分積分 >

integral from 1 to e-1 of 2xln(x+1)

解

\int_{1}^{e - 1} 2 x ln (x + 1) d x

解

2 (\frac{e ^{2}}{4} - 1)

+1

十進法表記

1.69452 \dots

解答ステップ

\int_{1}^{e - 1} 2 x ln (x + 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_{1}^{e - 1} x ln (x + 1) d x

u置換積分法を適用する

= 2 \cdot \int_{2}^{e} (u - 1) ln (u) d u

部分積分を適用する

= 2 [ln (u) (\frac{u ^{2}}{2} - u) - \int \frac{1}{2} (u - 2) d u]_{2}^{e}

\int \frac{1}{2} (u - 2) d u = \frac{u ^{2}}{4} - u

= 2 [ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u)]_{2}^{e}

簡素化

2 [ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u)]_{2}^{e} : 2 [\frac{1}{4} (2 u^{2} ln (u) - 4 u ln (u) - u^{2} + 4 u)]_{2}^{e}

= 2 [\frac{1}{4} (2 u^{2} ln (u) - 4 u ln (u) - u^{2} + 4 u)]_{2}^{e}

限度を計算する:

\frac{e ^{2}}{4} - 1

= 2 (\frac{e ^{2}}{4} - 1)

グラフ

人気の例

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