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人気のある微分積分 >

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

解

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

解

\frac{1}{2}

+1

十進法表記

0.5

解答ステップ

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

定数を除く:

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

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

u置換積分法を適用する

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

定数を除く:

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

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

べき乗則を適用する

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

簡素化

2 \cdot \frac{1}{2} [- \frac{1}{u}]_{1}^{2} : [- \frac{1}{u}]_{1}^{2}

= [- \frac{1}{u}]_{1}^{2}

限度を計算する:

\frac{1}{2}

= \frac{1}{2}

グラフ

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