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人気のある微分積分 >

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

解

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

解

\frac{6}{5}

+1

十進法表記

1.2

解答ステップ

\int_{0}^{2} \frac{3 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

= 3 \cdot \int_{0}^{2} \frac{x}{( x ^{2} + 1 ) ^{2}} d x

u置換積分法を適用する

= 3 \cdot \int_{1}^{5} \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

= 3 \cdot \frac{1}{2} \cdot \int_{1}^{5} \frac{1}{u ^{2}} d u

べき乗則を適用する

= 3 \cdot \frac{1}{2} [- \frac{1}{u}]_{1}^{5}

簡素化

3 \cdot \frac{1}{2} [- \frac{1}{u}]_{1}^{5} : \frac{3}{2} [- \frac{1}{u}]_{1}^{5}

= \frac{3}{2} [- \frac{1}{u}]_{1}^{5}

限度を計算する:

\frac{4}{5}

= \frac{3}{2} \cdot \frac{4}{5}

簡素化

= \frac{6}{5}

グラフ

人気の例

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