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人気のある微分積分 >

integral of (8x^2)/((64+x^2)^2)

解

\int \frac{8 x ^{2}}{( 64 + x ^{2} ) ^{2}} d x

解

\frac{1}{2} arctan (\frac{x}{8}) - \frac{4 x}{64 + x ^{2}} + C

解答ステップ

\int \frac{8 x ^{2}}{( 64 + x ^{2} ) ^{2}} d x

定数を除く:

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

= 8 \cdot \int \frac{x ^{2}}{( 64 + x ^{2} ) ^{2}} d x

置換積分法を適用する

= 8 \cdot \int \frac{u ^{2}}{8 ( 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

= 8 \cdot \frac{1}{8} \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}}

= 8 \cdot \frac{1}{8} \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

= 8 \cdot \frac{1}{8} (\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}})

= 8 \cdot \frac{1}{8} (arctan (u) - \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}}))

代用を戻す

u = \frac{x}{8}

= 8 \cdot \frac{1}{8} (arctan (\frac{x}{8}) - \frac{1}{2} (arctan (\frac{x}{8}) + \frac{\frac{x}{8}}{1 + ( \frac{x}{8} ) ^{2}}))

簡素化

8 \cdot \frac{1}{8} (arctan (\frac{x}{8}) - \frac{1}{2} (arctan (\frac{x}{8}) + \frac{\frac{x}{8}}{1 + ( \frac{x}{8} ) ^{2}})) : \frac{1}{2} arctan (\frac{x}{8}) - \frac{4 x}{64 + x ^{2}}

= \frac{1}{2} arctan (\frac{x}{8}) - \frac{4 x}{64 + x ^{2}}

定数を解答に追加する

= \frac{1}{2} arctan (\frac{x}{8}) - \frac{4 x}{64 + x ^{2}} + C

グラフ

人気の例

勾配 (2.6)(4.12)

s l o p e (2.6) (4.12)

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\int_{1}^{e^{6}} \frac{ln ^{2} ( x ^{2} )}{x} d x

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a re a x = 25 - y - 5, x = y - 10

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