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人気のある微分積分 >

integral from 0 to 1/2 of x/(2+32x^4)

解

\int_{0}^{\frac{1}{2}} \frac{x}{2 + 32 x ^{4}} d x

解

\frac{2 π - 4 arctan ( 2 + 1 ) + 4 arctan ( 2 - 1 )}{64}

+1

十進法表記

0.04908 \dots

解答ステップ

\int_{0}^{\frac{1}{2}} \frac{x}{2 + 32 x ^{4}} d x

\frac{x}{2 + 32 x ^{4}} = \frac{1}{2} \frac{x}{1 + 16 x ^{4}}

= \int_{0}^{\frac{1}{2}} \frac{1}{2} \cdot \frac{x}{1 + 16 x ^{4}} d x

定数を除く:

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

= \frac{1}{2} \cdot \int_{0}^{\frac{1}{2}} \frac{x}{1 + 16 x ^{4}} d x

以下の部分分数を得る:

\frac{x}{1 + 16 x ^{4}} : - \frac{1}{4 2 ( 4 x ^{2} + 2 2 x + 1 )} + \frac{1}{4 2 ( 4 x ^{2} - 2 2 x + 1 )}

= \frac{1}{2} \cdot \int_{0}^{\frac{1}{2}} - \frac{1}{4 2 ( 4 x ^{2} + 2 2 x + 1 )} + \frac{1}{4 2 ( 4 x ^{2} - 2 2 x + 1 )} d x

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

= \frac{1}{2} (- \int_{0}^{\frac{1}{2}} \frac{1}{4 2 ( 4 x ^{2} + 2 2 x + 1 )} d x + \int_{0}^{\frac{1}{2}} \frac{1}{4 2 ( 4 x ^{2} - 2 2 x + 1 )} d x)

\int_{0}^{\frac{1}{2}} \frac{1}{4 2 ( 4 x ^{2} + 2 2 x + 1 )} d x = \frac{1}{8} (arctan (2 + 1) - \frac{π}{4})

\int_{0}^{\frac{1}{2}} \frac{1}{4 2 ( 4 x ^{2} - 2 2 x + 1 )} d x = \frac{1}{8} (arctan (2 - 1) + \frac{π}{4})

= \frac{1}{2} (- \frac{1}{8} (arctan (2 + 1) - \frac{π}{4}) + \frac{1}{8} (arctan (2 - 1) + \frac{π}{4}))

簡素化

\frac{1}{2} (- \frac{1}{8} (arctan (2 + 1) - \frac{π}{4}) + \frac{1}{8} (arctan (2 - 1) + \frac{π}{4})) : \frac{2 π - 4 arctan ( 2 + 1 ) + 4 arctan ( 2 - 1 )}{64}

= \frac{2 π - 4 arctan ( 2 + 1 ) + 4 arctan ( 2 - 1 )}{64}

グラフ

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