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人気のある微分積分 >

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

解

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

解

2 ln^{2} (2) - 2 ln (2) + \frac{3}{4}

+1

十進法表記

0.32461 \dots

解答ステップ

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

u置換積分法を適用する

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

部分積分を適用する

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

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

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

簡素化

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

限度を計算する:

2 ln^{2} (2) - 2 ln (2) + \frac{3}{4}

= 2 ln^{2} (2) - 2 ln (2) + \frac{3}{4}

グラフ

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