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人気のある微分積分 >

integral of x/(1+x^6)

解

\int \frac{x}{1 + x ^{6}} d x

解

\frac{1}{12} (2 ln x^{2} + 1 - ln \frac{4}{3} + \frac{4 x ^{4} - 4 x ^{2}}{3} + 23 arctan (\frac{2 x ^{2} - 1}{3})) + C

解答ステップ

\int \frac{x}{1 + x ^{6}} d x

因数

\frac{x}{1 + x ^{6}}

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

u置換積分法を適用する

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

定数を除く:

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

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

以下の部分分数を得る:

\frac{1}{( u + 1 ) ( u ^{2} - u + 1 )} : \frac{1}{3 ( u + 1 )} + \frac{- u + 2}{3 ( u ^{2} - u + 1 )}

= \frac{1}{2} \cdot \int \frac{1}{3 ( u + 1 )} + \frac{- u + 2}{3 ( u ^{2} - u + 1 )} 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

= \frac{1}{2} (\int \frac{1}{3 ( u + 1 )} d u + \int \frac{- u + 2}{3 ( u ^{2} - u + 1 )} d u)

\int \frac{1}{3 ( u + 1 )} d u = \frac{1}{3} ln ∣ u + 1 ∣

\int \frac{- u + 2}{3 ( u ^{2} - u + 1 )} d u = \frac{1}{6} (- ln \frac{4}{3} + \frac{4 u ^{2} - 4 u}{3} + 23 arctan (\frac{2 u - 1}{3}))

= \frac{1}{2} (\frac{1}{3} ln ∣ u + 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 u ^{2} - 4 u}{3} + 23 arctan (\frac{2 u - 1}{3})))

代用を戻す

u = x^{2}

= \frac{1}{2} (\frac{1}{3} ln x^{2} + 1 + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 ( x ^{2} ) ^{2} - 4 x ^{2}}{3} + 23 arctan (\frac{2 x ^{2} - 1}{3})))

簡素化

\frac{1}{2} (\frac{1}{3} ln x^{2} + 1 + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 ( x ^{2} ) ^{2} - 4 x ^{2}}{3} + 23 arctan (\frac{2 x ^{2} - 1}{3}))) : \frac{1}{12} (2 ln x^{2} + 1 - ln \frac{4}{3} + \frac{4 x ^{4} - 4 x ^{2}}{3} + 23 arctan (\frac{2 x ^{2} - 1}{3}))

= \frac{1}{12} (2 ln x^{2} + 1 - ln \frac{4}{3} + \frac{4 x ^{4} - 4 x ^{2}}{3} + 23 arctan (\frac{2 x ^{2} - 1}{3}))

定数を解答に追加する

= \frac{1}{12} (2 ln x^{2} + 1 - ln \frac{4}{3} + \frac{4 x ^{4} - 4 x ^{2}}{3} + 23 arctan (\frac{2 x ^{2} - 1}{3})) + C

グラフ

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