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人気のある微分積分 >

integral of 24x^5arctan(2x^3)

解

\int 24 x^{5} arctan (2 x^{3}) d x

解

4 x^{6} arctan (2 x^{3}) - 2 x^{3} + arctan (2 x^{3}) + C

解答ステップ

\int 24 x^{5} arctan (2 x^{3}) d x

定数を除く:

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

= 24 \cdot \int x^{5} arctan (2 x^{3}) d x

u置換積分法を適用する

= 24 \cdot \int \frac{u arctan ( u )}{12} d u

定数を除く:

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

= 24 \cdot \frac{1}{12} \cdot \int u arctan (u) d u

部分積分を適用する

= 24 \cdot \frac{1}{12} (\frac{1}{2} u^{2} arctan (u) - \int \frac{u ^{2}}{2 ( u ^{2} + 1 )} d u)

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

= 24 \cdot \frac{1}{12} (\frac{1}{2} u^{2} arctan (u) - \frac{1}{2} (- arctan (u) + u))

代用を戻す

u = 2 x^{3}

= 24 \cdot \frac{1}{12} (\frac{1}{2} (2 x^{3})^{2} arctan (2 x^{3}) - \frac{1}{2} (- arctan (2 x^{3}) + 2 x^{3}))

簡素化

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

= 4 x^{6} arctan (2 x^{3}) - 2 x^{3} + arctan (2 x^{3})

定数を解答に追加する

= 4 x^{6} arctan (2 x^{3}) - 2 x^{3} + arctan (2 x^{3}) + C

グラフ

人気の例

integral of sqrt(e^{5t)}

\int e^{5 t} d t

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\int \frac{cos ( x )}{sin ( x ) + 2} d x

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\int 1 + y^{2} + x y^{2} d x

integral of 2x^3+5x^2-2x+pi

\int 2 x^{3} + 5 x^{2} - 2 x + π d x

integral of sqrt(sin^3(x)-sin^5(x))

\int sin^{3} (x) - sin^{5} (x) d x