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人気のある微分積分 >

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

解

\int_{0}^{1} x^{2} ln (x + 1) d x

解

- \frac{5}{18} + \frac{2}{3} ln (2)

+1

十進法表記

0.18432 \dots

解答ステップ

\int_{0}^{1} x^{2} ln (x + 1) d x

u置換積分法を適用する

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

部分積分を適用する

= [ln (u) (\frac{u ^{3}}{3} - u^{2} + u) - \int \frac{1}{3} (u^{2} - 3 u + 3) d u]_{1}^{2}

\int \frac{1}{3} (u^{2} - 3 u + 3) d u = \frac{u ^{3}}{9} - \frac{u ^{2}}{2} + u

= [ln (u) (\frac{u ^{3}}{3} - u^{2} + u) - (\frac{u ^{3}}{9} - \frac{u ^{2}}{2} + u)]_{1}^{2}

簡素化

[ln (u) (\frac{u ^{3}}{3} - u^{2} + u) - (\frac{u ^{3}}{9} - \frac{u ^{2}}{2} + u)]_{1}^{2} : [\frac{1}{3} u^{3} ln (u) - u^{2} ln (u) + u ln (u) - \frac{u ^{3}}{9} + \frac{u ^{2}}{2} - u]_{1}^{2}

= [\frac{1}{3} u^{3} ln (u) - u^{2} ln (u) + u ln (u) - \frac{u ^{3}}{9} + \frac{u ^{2}}{2} - u]_{1}^{2}

限度を計算する:

- \frac{5}{18} + \frac{2}{3} ln (2)

= - \frac{5}{18} + \frac{2}{3} ln (2)

グラフ

人気の例

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

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