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人気のある微分積分 >

integral of (2x^2)/((x^2+5)^2)

解

\int \frac{2 x ^{2}}{( x ^{2} + 5 ) ^{2}} d x

解

\frac{1}{5} arctan (\frac{x}{5}) - \frac{x}{5 + x ^{2}} + C

解答ステップ

\int \frac{2 x ^{2}}{( x ^{2} + 5 ) ^{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 \frac{x ^{2}}{( x ^{2} + 5 ) ^{2}} d x

置換積分法を適用する

= 2 \cdot \int \frac{u ^{2}}{5 ( u ^{2} + 1 ) ^{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}{5} \cdot \int \frac{u ^{2}}{( u ^{2} + 1 ) ^{2}} d u

以下の部分分数を得る:

\frac{u ^{2}}{( u ^{2} + 1 ) ^{2}} : \frac{1}{u ^{2} + 1} - \frac{1}{( u ^{2} + 1 ) ^{2}}

= 2 \cdot \frac{1}{5} \cdot \int \frac{1}{u ^{2} + 1} - \frac{1}{( u ^{2} + 1 ) ^{2}} d u

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \cdot \frac{1}{5} (\int \frac{1}{u ^{2} + 1} d u - \int \frac{1}{( u ^{2} + 1 ) ^{2}} d u)

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

\int \frac{1}{( u ^{2} + 1 ) ^{2}} d u = \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}})

= 2 \cdot \frac{1}{5} (arctan (u) - \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}}))

代用を戻す

u = \frac{x}{5}

= 2 \cdot \frac{1}{5} arctan (\frac{x}{5}) - \frac{1}{2} arctan (\frac{x}{5}) + \frac{\frac{x}{5}}{1 + ( \frac{x}{5} ) ^{2}}

簡素化

2 \cdot \frac{1}{5} arctan (\frac{x}{5}) - \frac{1}{2} arctan (\frac{x}{5}) + \frac{\frac{x}{5}}{1 + ( \frac{x}{5} ) ^{2}} : \frac{1}{5} arctan (\frac{x}{5}) - \frac{x}{5 + x ^{2}}

= \frac{1}{5} arctan (\frac{x}{5}) - \frac{x}{5 + x ^{2}}

定数を解答に追加する

= \frac{1}{5} arctan (\frac{x}{5}) - \frac{x}{5 + x ^{2}} + C

グラフ

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