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人気のある微分積分 >

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

解

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

解

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

+1

十進法表記

1.64791 \dots

解答ステップ

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

u置換積分法を適用する

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

部分積分を適用する

= [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

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

簡素化

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

= [\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{3}{2} ln (3)

グラフ

人気の例

integral from-infinity to 0 of x5^{-x}

\int_{- \infty}^{0} x 5^{- x} d x

integral from 0 to 1 of 1/(x^3+1)

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

integral from 1 to 2 of (x^2+4)/x

\int_{1}^{2} \frac{x ^{2} + 4}{x} d x

integral from 0 to 7/12 of 20x

\int_{0}^{\frac{7}{12}} 20 x d x

integral from 0 to infinity of 3/(x^2+1)

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