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人気のある微分積分 >

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

解

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

解

\frac{1}{4} (2 ln (2) - 1)

+1

十進法表記

0.09657 \dots

解答ステップ

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

簡素化

ln (x^{2} + 1) : \frac{1}{2} ln (x^{2} + 1)

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

定数を除く:

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

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

u置換積分法を適用する

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

定数を除く:

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

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

部分積分を適用する

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

\int 1 d u = u

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

簡素化

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

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

限度を計算する:

2 ln (2) - 1

= \frac{1}{4} (2 ln (2) - 1)

グラフ

人気の例

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\int_{0}^{1} \frac{1}{2} cosh (2 x) d x

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