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人気のある微分積分 >

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

解

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

解

\frac{1}{2} (5 ln (5) - 4)

+1

十進法表記

2.02359 \dots

解答ステップ

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

u置換積分法を適用する

= \int_{1}^{5} \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 \int_{1}^{5} ln (u) d u

部分積分を適用する

= \frac{1}{2} [u ln (u) - \int 1 d u]_{1}^{5}

\int 1 d u = u

= \frac{1}{2} [u ln (u) - u]_{1}^{5}

限度を計算する:

5 ln (5) - 4

= \frac{1}{2} (5 ln (5) - 4)

グラフ

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