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人気のある微分積分 >

integral of (6x^2)/((x^6+1)(x^3+1))

解

\int \frac{6 x ^{2}}{( x ^{6} + 1 ) ( x ^{3} + 1 )} d x

解

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

解答ステップ

\int \frac{6 x ^{2}}{( x ^{6} + 1 ) ( x ^{3} + 1 )} d x

定数を除く:

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

= 6 \cdot \int \frac{x ^{2}}{( x ^{6} + 1 ) ( x ^{3} + 1 )} d x

u置換積分法を適用する

= 6 \cdot \int \frac{1}{3 ( u ^{2} + 1 ) ( u + 1 )} d u

定数を除く:

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

= 6 \cdot \frac{1}{3} \cdot \int \frac{1}{( u ^{2} + 1 ) ( u + 1 )} d u

以下の部分分数を得る:

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

= 6 \cdot \frac{1}{3} \cdot \int \frac{- u + 1}{2 ( u ^{2} + 1 )} + \frac{1}{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

= 6 \cdot \frac{1}{3} (\int \frac{- u + 1}{2 ( u ^{2} + 1 )} d u + \int \frac{1}{2 ( u + 1 )} d u)

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

\int \frac{1}{2 ( u + 1 )} d u = \frac{1}{2} ln ∣ u + 1 ∣

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

代用を戻す

u = x^{3}

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

簡素化

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

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

定数を解答に追加する

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

グラフ

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