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人気のある微分積分 >

integral from 0 to 1 of xln(2x+1)

解

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

解

\frac{3}{8} ln (3)

+1

十進法表記

0.41197 \dots

解答ステップ

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

u置換積分法を適用する

= \int_{1}^{3} \frac{ln ( u ) ( u - 1 )}{4} d u

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int_{1}^{3} ln (u) (u - 1) d u

部分積分を適用する

= \frac{1}{4} [ln (u) (\frac{u ^{2}}{2} - u) - \int \frac{1}{2} (u - 2) d u]_{1}^{3}

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

= \frac{1}{4} [ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u)]_{1}^{3}

簡素化

\frac{1}{4} [ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u)]_{1}^{3} : \frac{1}{4} [\frac{1}{2} u^{2} ln (u) - u ln (u) - \frac{u ^{2}}{4} + u]_{1}^{3}

= \frac{1}{4} [\frac{1}{2} u^{2} ln (u) - u ln (u) - \frac{u ^{2}}{4} + u]_{1}^{3}

限度を計算する:

\frac{3}{2} ln (3)

= \frac{1}{4} \cdot \frac{3}{2} ln (3)

簡素化

= \frac{3}{8} ln (3)

グラフ

人気の例

integral from 5 to 9 of 1/(sqrt(9-x))

\int_{5}^{9} \frac{1}{9 - x} d x

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\int_{0}^{2 π} cos^{5} (x) d x

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\int_{0}^{1} \frac{1}{x ^{2} + 2 x + 5} d x

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